Selected chapters of geometry, analysis and number theory

SELECTED CHAPTERS OF GEOMETRY, ANALYSIS AND NUMBER THEORY JÓZSEF SÁNDOR ”... There is no study in the world which bri...

Author: József Sándor

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SELECTED CHAPTERS OF GEOMETRY, ANALYSIS AND NUMBER THEORY

JÓZSEF SÁNDOR

”... There is no study in the world which brings into more harmonious action all the faculties of the mind than mathematics... or like this, seems to raise them, by successive steps of initiation, to higher and higher states of conscious intellectual being...” (James J. Sylvester)

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Preface The aim of this book is to present short notes or articles, as well as studies on some topics of Geometry, Analysis, and Number Theory. The material is divided into ten chapters: Geometry and geometric inequalities; Sequences and series of real numbers; Special numbers and sequences of integers; Algebraic and analytic inequalities; Euler gamma function; Means and mean value theorems; Functional equations and inequalities; Diophantine equations; Arithmetic functions; and Miscellaneous themes. Chapter 1 deals essentially with classical geometric properties of triangles, polygons and tetrahedrons, as well as trigonometric functions. There are included some recent advances in triangle and tetrahedron inequalities, due to the author. Chapter 2 studies certain sequences and series of real numbers. Paragraph 2 includes some strange sequences, having interesting definitions and properties. Other themes introduce the Wallis product, Olivier’s criterion, Dirichlet’s beta function, or Bereznai’s theorem on the convergence or divergence of infinite series. The special numbers and sequences of integers of Chapter 3 require special methods of Number theory. Here some properties of the sequence of primes, of composite numbers, Bernoulli numbers, perfect numbers and generalizations, or abundant and deficient numbers, are studied. Chapter 4 discusses various algebraic and analytic inequalities, as the Cauchy-Bunjakovski, Chrystal’s, Hadamard’s, Jensen’s, Chebyshev’s, Fink’s, Ky Fan’s, Alzer’s, etc. inequalities, with many connections and applications to other fields. Chapter 5 contains 16 notes on the famous Euler gamma function. Many basic as well as new properties (due to the author), including monotonicity or convexity are studied. Papers 11-13 have been used by many authors in various fields of study, including Discrete mathematics, Number theory, Analysis, Linear Algebra, etc. Chapter 6 studies certain new mean value theorems, as well as mean values, with applications. A special attention is given to applications of Cauchy’s mean value theorem, as well as certain special means, including the so called logarithmic and identric means, or the Seiffert means. The first theme of Chapter 7 on the BohrMollerup theorem could have been considered also in Chapter 5 on Euler’s gamma function. Here the emphasis is however on functional equations, and along with the classical gamma function, the q-gamma function of Jackson is also involved. The other themes of this chapter rely on functional equations of 3

more unknowns as Cauchy’s, Haruki’s and Cior˘anescu’s, Rassias’ or Bartha’s functional equations. Some functional inequalities are studied, too. Chapter 8 contains certain Diophantine equations of a new type. The Pell type equations, as well as cubic equations of many unknowns are included. There are also other interesting equations involving factorials (based on a problem by Smarandache), or arithmetic functions ϕ, σ, ψ. Paragraph 10 introduces a new concept of Geometry and Number theory: harmonic triangles. A large chapter involving various arithmetic functions is Chapter 9. Here many properties of the classical functions as Euler’s totient, the sum and number- of divisors, Jordan’s totient, the Smarandache function, etc. are introduced. There are considered also many new arithmetic functions, as the Euler minimum and maximum functions, the Smarandache minimum and maximum functions, the star function of an arithmetic function, etc. Finally, Chapter 10 deals with various miscellaneous themes, as Fermat’s ”little” theorem and its generalizations, Euler pretty numbers, or certain inequalities of Klamkin, Ky Fan, or Lehman’s, with applications. Article 2 introduces the geometric numbers, not published elsewhere. The majority of the included material is original, based on papers by the author. Some of them have been published in some little known journals of Romania and Hungary, or in known journals from Bulgaria, Yugoslavia, Germany, Suisse, India, Australia, and USA. The book is concluded with an author index, focused on articles (and not pages), where the same author may appear more times. Finally, I wish to express my gratitude to a number of persons and organizations from whom I received valuable advice or support in the preparation of this book. These are the Mathematics and Informatics Departments of the Babe¸s-Bolyai University of Cluj (Romania), the Domus Hungarica Foundation of Budapest, the Sapientia Foundation and the Bolyai Society of Cluj; and also Professors H. Alzer, J. Aczél, T. Andreescu, K.T. Atanassov, M. Bencze, M. Balázs, W.W. Breckner, G.L. Cohen, P. Erdös, L. Filep, R.K. Guy, I. Kátai, J. Kolumbán, M.S. Klamkin, H.J. Kanold, A. Lupa¸s, R. Laatsch, Gy. Maurer, A. Makowski, E. Neuman, C. Pomerance, G. Robin, I.Z. Ruzsa, I. Ra¸sa, Th.M. Rassias, H.J.J. te Riele, K.B. Stolarsky, F. Smarandache, R.M. Sorli, M.V. Subbarao, H.N. Shapiro, A. Schinzel, H.J. Seiffert, L. Tóth, Gh. Toader, J. Wendel, I. Virág, M. Vuorinen, F.A. Valentine, A. Vernescu. The camera-ready manuscript for the present book was prepared by Mrs. Georgeta Bonda (Cluj) to whom the author expresses his gratitude. 16th August, 2005

The author

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Contents Preface

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1 Geometry and geometric inequalities 1 A property of polygons . . . . . . . . . . . . . . . . . . . 2 On a theorem of Cotes in elementary geometry . . . . . 3 On some new geometric inequalities . . . . . . . . . . . 4 Some inequalities for the elements of a triangle . . . . . 5 Recent advances in triangle inequalities . . . . . . . . . 6 The cotangent inequality of a triangle . . . . . . . . . . Y A Y A−B 7 On sin ≤ cos . . . . . . . . . . . . . . . 2 2 8 The sum of medians, and angle bysectors . . . . . . . . 9 On a generalization of the Erdös-Mordell inequality . . . 10 On certain inequalities in a tetrahedron . . . . . . . . . 11 Certain trigonometric inequalities deduced by convexity methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 On some trigonometric inequalities of Bencze . . . . . . sin x 13 On . . . . . . . . . . . . . . . . . . . . . . . . . . . x 14 On de Cusa’s trigonometric inequality . . . . . . . . . . 1 15 A property of sin . . . . . . . . . . . . . . . . . . . . . x 16 (sin x)cos x + (cos x)sin x made integer . . . . . . . . . . . 2 Sequences and series of real numbers 1 On certain limits related to the number e . . . . . . . . 2 On some strange sequences . . . . . . . . . . . . . . . . 3 Determination of certain sequences . . . . . . . . . . . . 4 A sequence connected with the Wallis product . . . . . 5 A generalized ratio test for series of positive terms . . . 6 A generalization of Bereznai’s theorem on infinite series 5

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11 12 13 15 19 22 27

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31 32 34 35

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40 49

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52 53

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54

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55

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57 58 64 90 92 94 95

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7 8 9 10

The iteration of sin and cos and two infinite series . . On Olivier’s criterion . . . . . . . . . . . . . . . . . . . On Dirichlet’s beta function . . . . . . . . . . . . . . . On sequences related to the sum of powers of positive integers . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . 98 . . . . . 99 . . . . . 101 . . . . . 102

3 Special numbers and sequences of integers 1 On pn / ln n . . . . . . . . . . . . . . . . . . . . . . . . . . √ 2 On the integer part of n pk . . . . . . . . . . . . . . . . . 3 [ex ] and [ey ] as consecutive primes . . . . . . . . . . . . . 4 On a sequence connected with the number of primes . . . 5 The smallest number of ones . . . . . . . . . . . . . . . . 6 On the sequence of composite numbers . . . . . . . . . . . 7 A note on Bernoulli numbers . . . . . . . . . . . . . . . . 8 The number of factorizations of n . . . . . . . . . . . . . . 9 On perfect and m-perfect numbers . . . . . . . . . . . . . 10 On Ruzsa’s lovely pairs . . . . . . . . . . . . . . . . . . . 11 Completely d-perfect numbers . . . . . . . . . . . . . . . . 12 On completely f -perfect numbers . . . . . . . . . . . . . . 13 On (f, g)-perfect numbers . . . . . . . . . . . . . . . . . . 14 On abundant and deficient numbers . . . . . . . . . . . . 15 On abundant and nobly-abundant, or deficient numbers . 16 The numbers n for which σ(n), d(n), ϕ(n) are abundant or deficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 On S-abundant and deficient numbers . . . . . . . . . . . 18 The square deficiency of a number . . . . . . . . . . . . . 19 Exponentially harmonic numbers . . . . . . . . . . . . . . 4 Algebraic and analytic inequalities 1 A monotonicity property of a product of sums of powers . 2 An application of the Cauchy-Bunjakovski inequality . . . 3 The Chrystal inequality and its applications to convexity 4 On certain inequalities for square roots and n-th roots . . k X √ n 5 On evaluation of i . . . . . . . . . . . . . . . . . . . 6 7 8 9

i=1

The product of consecutive odd integers On Hadamard’s inequality . . . . . . . . On certain inequalities for the number e The Jensen integral inequality . . . . . . 6

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105 106 107 108 108 109 110 112 113 114 116 118 121 126 130 132

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134 137 139 141

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147 148 148 150 152

. . . 155 . . . .

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157 159 163 165

10 11 12 13 14 15 16

Generalizations of certain integral inequalities The Chebyshev integral inequality . . . . . . On Fink’s inequality . . . . . . . . . . . . . . An extension of Ky Fan’s inequalities . . . . . A converse of Ky Fan’s inequality . . . . . . . A refinement of Gn + G0n ≤ 1 . . . . . . . . . On Alzer’s inequality . . . . . . . . . . . . . .

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5 Euler gamma function 1 A limit involving the Gamma function and the Lalescu sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 On a sequence containing the Gamma function . . . . . . 3 The product of consecutive factorials . . . . . . . . . . . . 4 On Γ(kn) . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 On a limit for the quotients of Gamma functions . . . . . 6 A limit for the Euler-Beta function . . . . . . . . . . . . . 7 On convex functions involving Euler’s Gamma function . 8 A convexity result on (f (x))1/x . . . . . . . . . . . . . . . 9 On a subadditive property of the Gamma function . . . . 10 A convexity property of the Gamma function . . . . . . . 11 Monotonicity and convexity (concavity) of some functions related to the Gamma function . . . . . . . . . . . . . . . 12 On the Gamma function II . . . . . . . . . . . . . . . . . 13 On the Gamma function III . . . . . . . . . . . . . . . . . 14 On certain inequalities for the ratios of Gamma functions 15 A note on certain inequalities for the Gamma function . . 16 Gamma function inequalities and traffic flow . . . . . . .

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169 175 178 179 182 183 184 187

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188 188 189 190 192 195 197 198 200 201

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202 210 214 221 224 226

6 Means and mean value theorems 1 Monotonicity and convexity properties of means . . . . . . . . 2 Logarithmic convexity of the means It and Lt . . . . . . . . . . 3 On certain subhomogeneous means . . . . . . . . . . . . . . . . 4 On a method of Steiner . . . . . . . . . . . . . . . . . . . . . . 5 On Karamata’s and Leach-Sholander’s theorems on means . . . 6 Certain logarithmic means . . . . . . . . . . . . . . . . . . . . . 7 The arithmetic-geometric mean of Gauss . . . . . . . . . . . . . 8 On two means by Seiffert . . . . . . . . . . . . . . . . . . . . . 9 Some new inequalities for means and convex functions . . . . . 10 On an inequality of Sierpinski on the arithmetic, geometric and harmonic means . . . . . . . . . . . . . . . . . . . . . . . . . .

229 230 237 240 248 250 251 256 265 268

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11 12 13 14 15 16

An application of Rolle’s theorem . . . . . . . . . . . . . . . . . An application of Lagrange’s mean value theorem for the computation of a limit . . . . . . . . . . . . . . . . . . . . . . . An inequality of Alzer, as an application of Cauchy’s mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A new mean value theorem . . . . . . . . . . . . . . . . . . . . Some mean value theorems and consequences . . . . . . . . . . The second mean value theorem of integral calculus . . . . . .

7 Functional equations and inequalities 1 The Bohr-Mollerup theorem . . . . . . . . . . . . . . . . . 2 On certain functional equations containing more unknown functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Locally integrable solutions of Cauchy’s functional equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Generalizations of Haruki’s and Cior˘anescu’s functional equations . . . . . . . . . Ã. . . . ! . . . . . . . . . . . . . . n n X X (f (xk ))k . . . . xk = 5 The functional equation f 6 7 8 9 10 11 12

k=1

k=1

Two functional equations . . . . . . . . . . . . . . . . A functional equation satisfied by the sin and identity functions . . . . . . . . . . . . . . . . . . . . . . . . . On certain functional equations by Rassias . . . . . . Bartha’s functional equation . . . . . . . . . . . . . . . An application of Chebyshev’s inequality . . . . . . . . On a functional inequality . . . . . . . . . . . . . . . . An unsolved functional equation . . . . . . . . . . . .

278 278 279 283 290

297 . . . 298 . . . 303 . . . 306 . . . 307 . . . 311

. . . . . 313 . . . . . .

8 Diophantine equations 1 Variations on a problem with factorials . . . . . . . . . . 2 On the Diophantine equation x1 x2 + x2 x3 + · · · + xn x1 = n + x1 + · · · + xn . . . . . . 3 On the equation n + (n + 1) + · · · + (n + k) = n(n + k) 4 On a problem of Subramanian, and Pell equations . . . 5 On the Diophantine equation x3 + y 3 + z 3 = a . . . . . 6 On the Diophantine equation x3 + y 3 + z 3 = a, II . . . . 7 n dimensional cuboids with integer sides and diagonals . 8 The equation {x2 } + {y 2 } = {z 2 } . . . . . . . . . . . . . 1 1 1 a 9 On the Diophantine equation + + ··· + = . x1 x2 xn b 8

274

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315 317 319 320 321 321

323 . . . . 324 . . . . . . .

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327 329 330 332 333 336 337

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10 11 12 13 14 15 16

Harmonic triangles . . . . . . . . . . . . . . . . . . . . . . . A Diophantine equation involving Euler’s totient . . . . . . On f (n) + f (n + 1) + · · · + f (n + k) = f (n)f (n + k) for f ∈ {ϕ, ψ, σ} . . . . . . . . . . . . . . . . . . . . . . . . On certain equations for the Euler, Dedekind, and Smarandache functions . . . . . . . . . . . . . . . . . . . . . Some equations involving the arithmetical functions ϕ, σ, ψ Equations with composite functions . . . . . . . . . . . . . On d(n) + σ(n) = 2n . . . . . . . . . . . . . . . . . . . . . .

9 Arithmetic functions 1 The non-Lipschitz property of certain arithmetic functions . 2 On certain open problems considered by Murthy . . . . . . 3 On an inequality of Moree on d(n) . . . . . . . . . . . . . . 4 On duals of the Smarandache simple function . . . . . . . . 5 A modification of the Smarandache function . . . . . . . . . 6 The star function of an arithmetic function . . . . . . . . . 7 On Jordan’s arithmetical function . . . . . . . . . . . . . . 8 Generalization of a theorem of Lucas on Euler’s totient . . . 9 Some arithmetic inequalities connected with the divisors of an integer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 A new arithmetic function . . . . . . . . . . . . . . . . . . . 11 A generalization of Ruzsa’s theorem . . . . . . . . . . . . . 12 On the monotonicity of the sequence (σk /σk∗ ) . . . . . . . . 13 A note on exponential divisors and related arithmetic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 The Euler minimum and maximum functions . . . . . . . . 15 The Smarandache minimum and maximum functions . . . . 16 The pseudo-Smarandache minimum and maximum functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Miscellaneous themes 1 On a divisibility problem . . . . . . . . . . . 2 A generalization of Fermat’s little theorem . . 3 On an inequality of Klamkin . . . . . . . . . 4 Euler-pretty numbers . . . . . . . . . . . . . . 5 Abundant numbers involving the smallest and factors . . . . . . . . . . . . . . . . . . . . . . 6 An inequality with sh x and ch x . . . . . . . 7 On geometric numbers . . . . . . . . . . . . . 9

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . largest prime . . . . . . . . . . . . . . . . . . . . . . . .

. . 341 . . 343 . . 346 . . . .

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350 350 351 353

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357 358 359 361 363 366 369 372 377

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379 382 388 389

. . 393 . . 398 . . 403 . . 408 . . . .

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413 414 414 416 422

. . 423 . . 424 . . 425

8 9 10

The sum-of-divisors minimum and maximum functions . . . . . 431 On certain new means and their Ky Fan type inequalities . . . 438 On Lehman’s inequality and electrical networks . . . . . . . . . 446

Author Index

455

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Chapter 1

Geometry and geometric inequalities ”... Nothing is more important than to see the sources of invention which are, in my opinion more interesting than the inventions themselves.” (Gottfried Leibnitz)

”... As long as algebra and geometry have been separated, their progress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched together towards perfection.” (Joseph-L. Lagrange)

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1

A property of polygons

Let A1 A2 . . . An be a convex polygon in the plane and P an arbitrary point. We will study the equality n X (−1)k P A2k = 0.

(1)

k=1

First, we show that n must be even. Indeed, let the coordinates of Ak (ak , bk ), while P (x, y). Then (1) is written equivalently as (x − a1 )2 + (y − b1 )2 + (x − a3 )2 + (y − b3 )2 + · · · = = (x − a2 )2 + (y − b2 )2 + (x − a4 )2 + (y − b4 )2 + . . . If n would be odd, then this equation would be the equation of circle, and clearly if P is not on this circle, relation (1) cannot be satisfied. Let n = 2k be even. Then the above equation gives: −2x(a1 + a3 + · · · + a2k−1 ) − 2y(b1 + b3 + · · · + b2k−1 ) + a21 + a23 + · · · + a22k−1 +b21 + · · · + b22k−1 = −2x(a2 + a4 + · · · + a2k ) − 2y(b2 + b4 + · · · + b2k ) +a22 + a24 + · · · + a22k + b22 + b24 + · · · + b22k . This clearly implies  a1 + a3 + · · · + a2k−1 = a2 + a4 + · · · + a2k    b1 + b3 + · · · + b2k−1 = b2 + b4 + · · · + b2k a2 + a2 + · · · + a22k−1 + b21 + b23 + · · · + b22k−1 =    1 2 3 2 = a2 + a4 + · · · + a22k + b22 + b24 + · · · + b22k

(2)

Therefore, conditions (2) are necessary and sufficient that (1) hold for arbitrary (fixed) points Ak . However, for a more transparent geometric study, we shall use the vectorial method. Let O 6= P another point with property (1). Then, since −−→ −−→ −−→ P Ai = P O + OAi , we get

i = 1, 2k,

−−→ −−→ P A2i = P O2 + OA2i + 2P O · OAi ,

giving: P A21 + P A23 + · · · + P A22k−1 − (P A22 + P A24 + · · · + P A22k ) 12

= OA21 + OA23 + · · · + OA22k−1 − (OA22 + OA24 + · · · + OA22k ) −−→ −−→ −−→ −−−−−→ −−→ −−→ −−−→ +2P O[(OA1 + OA3 + · · · + OA2k−1 ) − (OA2 + OA4 + · · · + OA2k )]. Since P and O satisfy (1) and P 6= O, we get: −−→ −−→ −−−−−→ −−→ −−→ −−−→ OA1 + OA3 + · · · + OA2k−1 = OA2 + OA4 + · · · + OA2k .

(3)

Let G1 , G2 be the centroids of polygons A1 A3 . . . A2k−1 and A2 A4 . . . A2k , respectively. Since −−→ −−→ −−−−−→ −−→ −−−→ −−−→ −−−−−−→ OA1 + OA3 + · · · + OA2k−1 = k OG1 + G1 A1 + G1 A3 + · · · + G1 A2k−1 and −−→ −−→ −−−→ −−→ −−−→ −−−→ −−−−→ OA2 + OA4 + · · · + OA2k = k OG2 + G2 A2 + G2 A4 + · · · + G2 A2k and

−−−→ −−−−−−→ −−−→ −−−−→ G1 A1 + · · · + G1 A2k−1 = G2 A2 + · · · + G2 A2k = 0, −−→ −−→ −−−→ we get OG1 = OG2 , i.e. G1 G2 = 0, implying G1 ≡ G2 . Therefore (3) means that the centroids of the above polygons must coincide. For example, when n = 4 (k = 2), we get a parallelogram, and condition OA21 +OA23 = OA22 +OA24 yields that this must be a rectangle. When n = 6 (k = 3), an example for which (1) is true is the regular hexagon, since then we may take O to be the centre of circumscribed circle when all conditions are satisfied.

2

On a theorem of Cotes in elementary geometry

Let (C) be a circle with centre O and radius r; let A0 , A1 , A2 , . . . , A2m−1 , A2m ≡ A0 be points on the circumference dividing it into 2m equal parts. If K is an arbitrary point of the diameter A0 Am , then R. Cotes (1682-1716) proved the following formulae: KA1 · KA3 . . . KA2m−1 = rm ± xm , KA0 · KA2 . . . KA2m = rm ± xm , where x = OK, and the sign depends on the parity of m and the position of K to the point O (see also [1]). In what follows, the following generalization will be deduced. 13

Theorem. Let (X, +, ·) be a commutative field and let (X ∗ , ·) be the multiplicative subgroup of X. Let G = {ai : i = 1, 2, . . . , n} be a finite subgroup of X ∗ . If 1 ∈ G is the unity element of G, then the following identity holds true: n Y (x + tai ) = xn + (−1)n−1 tn ,

x, t ∈ X arbitrary.

i=1

Proof. We will apply Fermat’s theorem for finite groups: If the order of the finite group G is N , then g N = 1 for all g ∈ G (see [2]). Let us consider the polynomial P (u) = un + (−1)n−1 tn (u ∈ X). We’ll prove that xi = −tai (i = 1, 2, . . . , n) are roots for P . Indeed, by using the properties of field X and the above theorem of Fermat, one has: P (xi ) = (−tai )n + (−1)n−1 tn = (−1)n tn ani + (−1)n−1 tn = tn [(−1)n + (−1)n−1 ] = tn · 0 = 0, where 0 is the neutral element of X. Now, Bézout’s theorem implies the stated identity. We now prove that Cotes’ theorem is a consequence of this result. Let the affixes of the points Ak in the complex plane be Ak (rekiπ/m ) (where i2 = −1) and of K be a K(a); and let us consider the set G of numbers a0 = ei·0π/m , a2 = ei2π/m , . . . , a2m−2 = ei(2m−2)π/m . It is immediate that (G, ·) is a finite subgroup of the multiplicative group (C ∗ , ·) of nonzero complex numbers (indeed: as at ∈ G, a−1 s = as , s, t = 0, . . . , m − 2). By the Theorem one has n−1 Y

KA2k = |am + (−1)m (−r)m |.

k=0

We have two situations: If K and A0 are on the same part with respect to O, when a > 0, a = x and r > x, so the result is rm − xm , in the second case K and A0 are in distinct sides of the diameter with respect to O, then a = −x, so if m is even the result is rm − xm ; but rm + xm if m is odd. Applying the theorem to the finite group of roots of unity of order 2m we get: 2m−1 Y KAk = |r2m − a2m | = r2m − x2m k=0

(since 2m is even). Therefore m Y k=1

KA2k−1 =

r2m − x2m r2m − x2m = = r m + xm m−1 rm ± xm Y KA2k k=0

14

in case 1), while in case 2) we get rm + xm for m even, rm − xm for m odd.

References [1] The Pallas Great Lexicon (Hungarian), Budapest, 1983, IV, 542. [2] Gy. Maurer, I. Virág, Introduction to structure theory (Hungarian), Ed. Dacia, Cluj, 1976.

3

On some new geometric inequalities

1. Let ABC be a triangle with standard notations, see e.g. our paper [3]. Let ha , hb , hc be the altitudes; ma , mb , mc the medians; and la , lb , lc the angle bisectors of the triangle. Let T = area(ABC). In the book [2] (Result 2.12) it is proved the following implication a ≥ b ⇒ a + ha ≥ b + hb .

(1)

Pál Erdös raised the problem ?

a > b > c ⇒ a + la > b + lb > c + lc , and Béla Finta [1] infirmed this property. Of course, remains open the problem of characterization of ? a ≥ b ⇒ a + la ≥ b + lb . (2) In this section we shall obtain certain results of type (1) and (2). As a generalization of (1), the following is true a ≥ b ⇒ aα + hαa ≥ bα + hαb

(3)

for all α ∈ R. Indeed, let α > 0. Then, since aha = bhb = 2T , the inequality to be proved becomes aα − bα ≥

(2T )α (aα − bα ) , (ab)α

which is equivalent to ab ≥ 2T . This is valid, since 2T = ab sin C ≤ ab. When −β + h−β is equivalent to α < 0, put α = −β, and remark that a−β + h−β a ≥b b hβa + aβ ≥ hβb + bβ (β > 0) which has been proved above. Thus (3) holds true for all real numbers α. 15

2. General situations, such in (3), are very rare, and difficult to obtain. For ma or la we shall consider mainly the case α = 2. First we prove another elementary, but nice fact a ≥ b ⇒ a2 + m2a ≥ b2 + m2b .

(4)

1 By m2a = [2(b2 + c2 ) − a2 ], etc., we have 4 a2 + m2a =

2b2 + 2c2 + 3a2 4

and b2 + m2b =

2a2 + 2c2 + 3b2 , 4

so 2b2 + 2c2 + 3a2 ≥ 2a2 + 2c2 + 3b2 ⇔ a2 ≥ b2 ⇔ a ≥ b. On the other hand, a multiplicative analog stand as follows: If a ≥ b, then ama ≥ bmb iff a2 + b2 ≤ 2c2

(5) (∗)

Indeed, a2 m2a ≥ b2 m2b ⇔

a2 b2 [2(b2 + c2 ) − a2 ] ≥ [2(a2 + c2 ) − b2 ] ⇔ 4 4

2a2 c2 − a4 ≥ 2b2 c2 − b4 ⇔ b4 − a4 + 2c2 (a2 − b2 ) ≥ 0 ⇔ (a2 − b2 )[2c2 − (a2 + b2 )] ≥ 0 ⇔ 2c2 ≥ a2 + b2 . b ≥ 90◦ , then inequality (∗) is true. Indeed, in this case it Remark. If m(C) √ is known the inequality a + b ≤ c 2 (see our monograph [2]). Thus a2 + b2 < (a + b)2 ≤ 2c2 . 3. A multiplicative analog involving la holds only when certain additional conditions are satisfied: b ≤ 60◦ or m(C) b ≥ 90◦ , then ala ≤ blb . If a ≥ b and m(C) (6) ◦ ◦ b ≤ 90 , then ala ≥ blb . If a ≥ b and 60 ≤ m(C) (7) 2 By using the classical formulae for la , we can write successively a2 la2 = iff

2a2 bcp(b + c − a) 2b2 acp(c + a − b) ≥ = b2 lb2 (b + c)2 (a + c)2 b(c + a − b) a(b + c − a) ≥ . 2 (b + c) (a + c)2 16

Let

b a , y= . a+c b+c Then the above inequality can be written as x=

y − y 2 ≥ x − x2 But

or (x − y)(x + y − 1) ≥ 0.

b a (b − a)(b + a + c) − = ≤0 a+c b+c (a + c)(b + c)

x−y =

(after certain elementary calculations) if a ≥ b. On the other hand x + y ≤ 1 can be written as b2 + bc + a2 + ac ≤ ab + ac + cb + c2

or a2 + b2 − ab ≤ c2 .

1 By using the law of cosines, c2 = a2 + b2 − 2ab cos C, thus cos C ≤ which 2 b ≤ 60◦ or m(C) b ≥ 90◦ . is true only if m(C) 4. We now remark that a ≥ b ⇒ a + ma ≥ m + mb

(8)

3 ma + mb ≥ (a + b). 4

(∗∗)

iff Indeed,

a + ma ≥ b + mb ⇔ 2a + or 2(a − b) ≥

p

2(b2 + c2 ) − a2 ≥ 2b +

p

2(a2 + c2 ) − b2

p p 2(a2 + c2 ) − b2 − 2(b2 + c2 ) − a2

[2(a2 + c2 ) − b2 ] − [2(b2 + c2 ) − a2 ] p =p 2(a2 + c2 ) − b2 + 2(b2 + c2 ) − a2 3(a2 − b2 ) p . 2(a2 + c2 ) − b2 + 2(b2 + c2 ) − a2

=p

By simplifying with a − b > 0, we get ma + mb ≥ 34 (a + b). Remarks. 1) it is easy to see that in all triangles ABC one has ma + mb ≥ 3c 3 3c 2 . Since 2 ≥ 4 (a + b) ⇔ a + b ≤ 2c, we get the following corollary (by taking into account of (∗∗)) If a ≥ b and a + b ≤ 2c, then a + ma ≥ b + mb . (9) 17

2) Let G be the centroid of the triangle ABC, and let A0 = AG ∩ BC, BDkCG, CDkBG. Then, it is easy to see, that in 4BGD: 2 BG = mb , 3

2 BD = GC = mc , 3

1 2 GD = 2 · ma = ma . 3 3

a+b (where C 0 is the midpoint of [AB]) then ma + 2 mb ≥ 43 (a + b) can be written as BG + GD ≥ C 0 A0 + BA0 . This must be true in the trapezium C 0 GDB. Clearly, this is not generally valid, so the inequality (8) itself is valid under additional conditions. 3) Another remark is that in 4GDB, from BG + GD > BD we get \ ≥ 90◦ , it is well known (see [2], p.47) ma + mb > mc . Assuming that m(BGD) √ √ a+b 3 2 and < , we get the following that ma + mb ≤ 2mc . Since mc < 2 2 4 corollary: \ ≥ 90◦ , and a ≥ b, then a + ma ≤ b + mb . If m(BGD) (10) 5. We now prove that Since C 0 A0 + BA0 =

a ≥ b ⇒ a2 + la2 ≥ b2 + lb2

(11)

iff c(a + b + c)(c3 + ba2 + ab2 + bc2 + ac2 + 3abc) ≤ (b + a)(a + c)2 (b + c)2 . (∗ ∗ ∗) Indeed, by using the formula la =

2 p bcp(p − a), b+c

one gets la2 + lb2 = (b − a)(c3 + ba2 + bc2 + ab2 + ac2 + 3abc)

c(a + b + c) (a + c)2 (b + c)2

≥ (b − a)(b − a) if (∗ ∗ ∗) is valid. Remark. Thus, the problem which arises here, is the validity of condition (∗ ∗ ∗). The analogous linear variant is the following a ≥ b ⇒ a + la ≥ b + lb iff (12) la + lb ≥ δ(a, b, c), 18

(∗ ∗ ∗∗)

where δ(a, b, c) = (c3 + 3abc + ba2 + ab2 + bc2 + ac2 )

a(a + b + c) . (a + c)2 (b + c)2

We have proved above that la2 − lb2 = (b − a)δ(a, b, c), thus la − lb =

(b − a)δ(a, b, c) ≥ b − a ⇔ δ(a, b, c) ≤ la + lb . la + lb

Problems. The √ study of expression δ(a, b, c). For example, since it is known that la +lb ≤ p 3−mc (see [2], p.49) one can ask, under what conditions do we have √ p 3 − mc ≤ δ?

√ One can ask, if p 3 − δ has a constant sign, or √ δ < p 3?

(13)

(14)

A similar question is the following one: if a ≥ b, under what conditions can be written the inequality δ ≤ a + b? Of course, one can raise problems for relations (4), etc. by asking conditions for a ≥ b ⇒ aα + mαa ≥ bα + mαb (α ∈ R, for example).

References al Erd¨ os, Oc[1] B. Finta, A solution for an elementary open question of P´ togon Math. Mag., 4(1996), no.1, 74-79. [2] J. Sándor, Geometric inequalities (Hungarian), Ed. Dacia, Cluj, 1988. [3] J. Sándor, On certain inequalities for the distances of a point to the vertices and the sides of a triangle, Octogon Math. Mag., 5(1997), no.1, 19-23.

4

Some inequalities for the elements of a triangle

In this section certain new inequalities for the angles (in radians) and other elements of a triangle are given. For such inequalities we quote the monographs [2] and [3]. 19

1. Let us consider the function f (x) =

x , sin x

0<x 0. sin x

Hence the function f is monotonous nondecreasing on (0, π), so that one can write f (B) ≤ f (A) for A ≤ B, i.e. B A ≤ , b a

(1)

b a because of sin B = and sin A = . Then, since B ≤ A if b ≤ a, (1) 2R 2R implies the relation a A ≥ , if a ≥ b. (i) B b 2. Assume, without loss of generality, that a ≥ b ≥ c. Then in view of (i), A B C ≥ ≥ , a b c and consequently µ ¶ A B (a − b) − ≥ 0, a b

µ (b − c)

B C − b c

¶

µ ≥ 0,

(c − a)

C A − c a

¶ ≥ 0.

Adding these inequalities, we obtain µ ¶ X A B (a − b) − ≥ 0, a b i.e.

X A (b + c) . a Adding A + B + C to both sides of this inequality, and by taking into account of A + B + C = π, and a + b + c = 2s (where s is the semi-perimeter of the triangle) we get XA 3π ≤ . (ii) a 2s This may be compared with Nedelcu’s inequality (see [3], p.212) XA 3π (ii)’ < . a 4R 2(A + B + C) ≥

20

Another inequality of Nedelcu says that X 1 2s > . (ii)” A πr Here r and R represent the radius of the incircle, respectively circumscribed circle of the triangle. 3. By the arithmetic-geometric inequality we have µ ¶1 XA ABC 3 ≥3 . (3) a abc Then, from (ii) and (2) one has ³ π ´3 ABC ≤ , abc 2s that is

µ ¶3 abc 2s ≥ . ABC π 4. Clearly, one has µ √ √ ¶2 µ √ √ ¶2 √ ¶2 µ √ y y x z z x √ √ √ √ √ √ − + − + − ≥ 0, b By a Ax b By c Cz a Ax c Cz (iii)

or equivalently, y + z bc z + x ca x + y ab · + · + · ≥2 x aA y bB z cC

µ

a b c √ +√ +√ BC CA AB

¶ . (3)

By using again the A.M.-G.M. inequality, we obtain µ ¶1 a b c abc 3 √ +√ +√ ≥3 . ABC BC CA AB Then, on base of (iii), one gets a b c 6s √ +√ +√ ≥ . π BC CA AB

(4)

Now (4) and (3) implies that y + z bc z + x ca x + y ab 12s (iv) · + · + · ≥ . x aA y bB z cC πµ ¶ 1 1 1 By putting (x, y, z) = (s − a, s − b, s − c) or , , in (iv), we can a b c deduce respectively bc ca ab 12s + + ≥ , A(s − a) B(s − b) C(s − c) π 21

b+c c+a a+b 12s + + ≥ , A B C π

which were proved in [1].

h πi 2 5. By applying Jordan’s inequality sin x ≥ x, x ∈ 0, , (see [3], p.201) π 2 in an acute-angled triangle, we can deduce, by using a = 2R sin A, etc. that Xa 12 (v) > R. A π By (ii) and the algebraic inequality µ ¶ 1 1 1 (x + y + z) + + ≥ 9, x y z

clearly, one can obtain the analogous relation (in every triangle) Xa 6 (v)’ ≥ s. A π Now, Redheffer’s inequality (see [3], p.228) says that sin x π 2 − x2 ≥ 2 for x ∈ (0, π). x π + x2 √ X 3 3 , an easy calculation yields the following interesting Since sin A ≤ 2 inequality √ X A3 3 3 (vi) . >π− π 2 + A2 4 Similarly, without using the inequality on the sum of sin’s one can deduce Xa X π 2 − A2 (vii) > 2R . A π 2 + A2 From this other corollaries are obtainable.

References ˇ Arslanagić, D.M. Milosević, Problem 1827, Crux Math., Canada, [1] S. 19(1993), 78. [2] D.S. Mitrinović et al., Recent advances in geometric inequalities, Kluwer Acad. Publ., 1989. [3] J. Sándor, Geometric inequalities (Hungarian), Ed. Dacia, Cluj, 1988.

5

Recent advances in triangle inequalities In what follows we will follow the terminology of [5]. Let ABC be a triangle. 1. Let t, λ be arbitrary real numbers. By Cauchy’s inequality, we can write: ³X ´ ³X ´ ³X ´ aλ−(t/2) at/2 wa ≤ a2λ−t at wa2 , 22

where the sums are cyclic. Since wa2 ≤ s(s − a) and

X

at (s − a) ≤

[1]), we obtain the inequality X

r

³X ´ 1 abc at−2 (see 1.10 from 2

abcs ³X 2λ−t ´ ³X t−2 ´ a a , 2

λ

a wa ≤

λ, t ∈ R.

(1)

This inequality generalizes some known results. For t = 2 one has 8.12 in [1], while for t = 3 we get relation (B) in [3]. The same technique can be used for ³X

an−1 a(t−1)/2 cos(α/2)

´2

≤

³X

a2n−2

´ ³X

´ at−1 cos2 (α/2) .

Since X

at−1 cos2 (α/2) =

X

at−1

´ s(s − a) s ³X t = a (s − a) , bc abc

by the quoted 1.10 inequality easily follows r ³ X s X 2n−2 ´ ³X t−2 ´ (2n+t−3)/2 a cos(α/2) ≤ a a . 2

(2)

Here n, t ∈ R are arbitrary real numbers. For t = 3 we get the result (4) from [2]. 2. It is well-known that wa2 ≤ s(s − a). In fact, since 1 1 m2a = (2b2 + 2c2 − a2 ) = s(s − a) + (b − c)2 ≥ s(s − a), 4 4 we have wa2 ≤ s(s − a) ≤ m2a ,

(3)

which refines wa ≤ ma . Now ³X p ´2 ³X √ ´2 ³X ´ as(s − a) ≤ ama ≤ 3 am2a 8 = 3 · (s2 + 2Rr + 5r2 ), 2 getting

³X p

´2 3 a(s − a) ≤ (s2 + 2Rr + 5r2 ). 2 23

(4)

We note that inequality is more precise than Xp √ a(s − a) ≤ s 2, i.e. inequality 5.47 in [5] (see Appendix, p.679). As a + b + c = 2s and a2 + b2 + c2 = 2(s2 − 4Rr − r2 ), we have X a(s − a) = 2r(4R + r), so

³X p ´2 ³X ´ a(s − a) ≤ 3 a(s − a) = 6r(4R + r),

(5)

improving relation (4). 3. It is well-known that ³α´ 2bc 2bc a · ctg (α/2) 1 cos · wa ≤ ≤ ≤ a · ctg (α/2), 2 b+c 2 b+c (b + c) 2 thus wa ≤ 2R cos2 (α/2). Because of

1 cos(α/2) ≥ (sin β + sin γ), 2

i.e. cos(α/2) ≥ one has also wa ≥

b+c 4R

bc . 2R

Consequently:

bc ≤ wa ≤ 2R cos2 (α/2). 2R For an application consider µX ¶ X 1 1 1 ≤ 2R = wa ab r and by

with the inequality

³X

´ µX 1 ¶ wa ≥ 9, wa X

wa ≤ 24

9R 2

(6)

it results:

X 1 2 ≥ . wa R

Thus:

X 1 2 1 ≤ , ≤ R wa r

(7)

which is a refinement of Euler’s classical inequality. For an application of the right-side of (6), we can use X √ a cos(α/2) ≤ s 3 (take e.g. n = 1, t = 3 in (2)). By √ √ a wa ≤ 2R · a cos(α/2) and the above relation it follows: X √ √ a wa ≤ s 6R. Inequality (7) has been discovered also in 1987 by M. Bencze, see e.g. M. Bencze, A refinement of the Euler’s inequality R ≥ 2r, Octogon Math. Mag. 5(1997), no.2, 39-47. 4. In paper [4], Theorem 11 we have proved that X at (s − a) ≥ s(4r(R + r))t/2 for t ≥ 2. Here

X

³X ´ 1 at (s − a) ≤ abc at−2 , 2

so have

³X ´ 1 s(4r(R + r))t/2 ≤ abc at−2 , 2 For t = 3 this gives:

t ≥ 2.

(abc)3 ≥ (4r(R + r))3 . µ This inequality is better that

(abc)2

[1].

≥

4R √ 3

(10)

¶3 , which appears as 4.14 in

5. Since 1 m2a = (2b2 + 2c2 − a2 ), 4 25

(9)

1 m2b = (2c2 + 2a2 − b2 ) 4

and we have

1 m2c = (2a2 + 2b2 − c2 ), 4 1 (−4(a6 + b6 + c6 ) + 3a2 b2 c2 64 +6(a4 b2 + b2 c4 + b4 c2 + a4 c2 + a2 c4 + a2 b4 )). m2a m2b m2c =

(11)

Then, by a4 b2 + a2 b4 ≤ a6 + b6 ,

b4 c2 + b2 c4 ≤ b6 + c6

and a4 c2 + a2 c4 ≤ a6 + c6 , (11) implies

´1/2 1 ³ ³X 6 ´ 8 a + 3a2 b2 c2 , 8 containing an extension of IX.10.12(a) in [5]. 6. It is well-known that √ 2 bc p s(s − a), wa = b+c so √ X wa X 2 bc p = s(s − a). s= a a(b + c) p √ By 2 bc ≤ b + c and a = (s − b) + (s − c) ≥ 2 (s − b)(s − c), from we obtain: p X s(s − a) p . s≤ 2 (s − b)(s − c) s Then, since (s − a)(s − b)(s − c) = r2 s, (14) yields S ≤ , that is, 2r X wa s ≤ . a 2r 7. On the basis of X s2 + (4R + r)2 sec2 (α/2) = s2 as well as inequality 5.7 in [1], i.e. ma mb mc ≤

(12)

(13) (13) (14)

(15)

2s2 (2R − r) ≤ R(4R + r)2 , we get

X

sec2 α/2 ≥ 5 −

sharpening GI 2.48 in [1]. See also [6]. 26

2r , R

(16)

References [1] O. Bottema and oth., Geometric inequalities, Wolters-Noordhoff, Gröningen, 1969. [2] D.M. Milosević, Some inequalities for a triangle, Punime Matematike, 2(1987), 23-30. [3] D.M. Milosević, Some inequalities for the triangle, Punime Matematike, 3(1988), 35-40. [4] D.M. Milosević, Recent advances in triangle inequalities, Punime Matematike, 5(1990), 35-40. [5] D.S. Mitrinović, J.E. Pecarić, V. Volenec, Recent advances in geometric inequalities, Kluwer, Dordrecht, 1989. [6] J. Sándor, D.M. Milosević, Recent advances in triangle inequalities, II, Octogon Math. Mag., 10(2002), no.2, 681-684.

6

The cotangent inequality of a triangle 1. Let ABC be a triangle. The well-known identity tg A + tg B + tg C = tg A · tg B · tg C

can be written with cotangents also as X ctg A · ctg B = 1. Now, using the classical inequality X X x2 ≥ xy, one can write i.e.

giving

X ³X

x, y ∈ R

ctg 2 A ≥ 1,

´2 X ctg A − 2 ctg A · ctg B ≥ 1, ³X

´2 ctg A ≥ 3, 27

i.e.

X

ctg A ≥

√ 3.

(1)

This is the classical cotangent inequality of a triangle, with many known proofs in the literature (for the above simple proof, see [4], p.102). Now, since a2 a2 sin B sin C = = 4R2 sin A sin B sin C ctg B + ctg C sin(B + C) = 4R2 we get

abc abc = = 2S, 3 8R 2R

a2 = 2S, ctg B + ctg C

(2)

where S is the area of triangle ABC. Writing two similar relations, and solving the obtained system of linear equations, one can deduce: ctg A =

b2 + c2 − a2 , 4S

ctg B =

a2 + c2 − b2 , 4S

ctg C =

b2 + a2 − c2 . 4S

(3)

For example, (3) gives the identity X

ctg A =

a2 + b2 + c2 4S

which, by (1) implies the famous inequality √ a2 + b2 + c2 ≥ 4 3 · S.

(4)

(5)

It seems that the first who proved (5) was R. Weitzenb¨ ock in 1919 (see [6]). For at least five distinct proofs, see also [4], [5]. We give here an apparently new proof of (5). This is based on the identity X X 16S 2 = 2 a2 b2 − a4 (6) which follows at once from Heron’s formula for the area of a triangle. Now, to prove (5) we have to verify that ³ X X ´ ³X ´2 X X 3 2 a2 b2 − a4 ≤ a2 = a4 + 2 a2 b2 . This is equivalent to

X

a2 b2 ≤ 28

X

a4 ,

so again a particular case of X X xy ≤ x2 ,

x, y, z ∈ R.

2. In order to obtain strong refinements of (5), apply first the cotangent inequality (1) to a triangle DEF having angles D=

π A − , 2 2

Then (1) gives:

E= X

π B − , 2 2

F =

π C − . 2 2

A √ ≥ 3. 2 This could be called as the ”tangent inequality” of a triangle. Now, tg

tg

(7)

1 − cos A 1 A = = − ctg A. 2 sin A sin A

Remark also that

1 bc = , sin A 2S so using relation (4), too; (7) will imply: X X √ 2 ab − a2 ≥ 4 3 · S.

(8)

By (5), this will give also X

√ ab ≥ 4 3 · S

(9)

which clearly improves (5). Therefore, we have deduced the following chain of improvements: X X X X √ (10) a2 ≥ ab ≥ 2 ab − a2 ≥ 4 3 · S. 3. Let R be the radius of the circumscribed circle. It is well-known that X √ (11) a ≤ 3R 3, so

4S abc 3abc √ = √ ≤ , a+b+c 3 R 3

by (11). Now, by the arithmetic-geometric inequality √ 3 a + b + c ≥ 3 abc, 29

this will give:

√ 3p 3 (abc)2 . (12) S≤ 4 Inequality (12) is due to Pólya and Szegö (who used differential calculus for proof). We note that Bottema et al. [1] attribute (12) to the later authors L. Carlitz and F. Leuenberger. Now, since X p ab ≥ 3 3 (abc)2 (by the arithmetic-geometric inequality), so by (12), we get the following chain of inequalities, similar to (10): X X p √ a2 ≥ ab ≥ 3 3 (abc)2 ≥ 3 3 · S. (13) Remark. The third terms in (10), respectively (13) cannot be compared to each other: Inequalities X X p (14) 2 ab − a2 ≥ 3 3 (abc)2 , 2

X

ab −

X

p a2 ≥ 3 3 (abc)2

(15)

are not valid in any triangle ABC. For example, (14) is not true for a = 2, 1 b = 2, c = 1; while inequality (15) is not true for e.g. a = 1, b = 1, c = . 2 4. Now, let U V W be another triangle, having as sides u, v, w and area S 0 . As an extension of the cotangent inequality (1), we will prove: X u2 · ctg A ≥ 4S 0 . (16) By (3) this can be written as u2 (−a2 + b2 + c2 ) + v 2 (a2 − b2 + c2 ) + w2 (a2 + b2 − c2 ) ≥ 16S · S 0 , discovered by D. Pedoe in 1942 (see [1]). ab uv Since S = sin x, S 0 = sin y, by 2 2 c2 = a2 + b2 − 2ab cos x,

w2 = u2 + v 2 − 2uv cos y,

the above relation can be written also as 4abuv sin x sin y ≤ u2 (2b2 − 2ab cos x) + v 2 (2a2 − 2ab cos x) 30

+(u2 + v 2 − 2uv cos y) · 2ab cos x, i.e. 4abuv(sin x sin y + cos x cos y) ≤ 2(u2 b2 + v 2 a2 ), i.e. (ub − va)2 + 2abuv[1 − cos(x − y)] ≥ 0 which is trivial, since cos(x − y) ≤ 1. One has equality only for ub = va, a b cos(x − y) = 0, i.e. = , x = y; i.e. when the two triangles are similar. u v When U V W is an exact triangle, having sides u = v = w = k, then √ k2 3 0 , S = 4 so (16) reduces to the cotangent inequality (1). 5. Finally we note that the cotangent inequality (1) is related also to the ”Brocard angles” of a triangle (see [3]), or to the ”Lemoine point” of a triangle (see e.g. [5]).

References [1] O. Bottema et. al., Geometric inequalities, Groningen, 1968. atze aus der Analysis, II, Leipzig, [2] G. Pólya, G. Szegö, Aufgaben und Lehs¨ 1925. [3] T. Lalescu, La géométrie du triangle, Paris, 1937 (Romanian), Ed. Tineretului, 1958. [4] J. Sándor, Geometric inequalities (Hungarian), Ed. Dacia, Cluj, 1988. [5] J. Sándor, On the Lemoine point of a triangle (Romanian), Lucr. Sem. Did. Mat. 16(2000), 175-182. [6] R. Weitzenböck, Math. Z., 5(1919), 137-146.

7

On

Y

A−B A Y cos sin ≤ 2 2

We will obtain the best α ≥ 0 such that in a triangle ABC holds true: 8

Y

sin

A Y A−B ≤ cos . 2 2+α 31

(1)

We will show that α = 0 is the best constant. By the Mollweide relations B−C b+c 2 = A a sin 2

cos

etc.

so Y

B−C √ √ √ cos b+c a+b a+c 2 bc 2 ab 2 ac 2 ≥ · · ≥ · · = 8, Y A a c b a c b sin 2

with equality in an equilateral triangle. Now remark that |B − C| |B − C| ≥ 2 2+α so cos giving

Y

|B − C| |B − C| ≥ cos 2+α 2

etc.,

Y |B − C| |B − C| cos cos 2+α ≥ 2 ≥ 8, Y Y A A sin sin 2 2

since in fact

|B − C| B−C = cos , etc. 2 2 Therefore the best α is α = 0. The inequality cos

8

Y

sin

A Y A−B ≤ cos 2 2

improves the well-known results that Y

8

sin

A 1 ≤ . 2 8

The sum of medians, and angle bysectors We have to determine all functions f : R2 → (1, ∞) such that ma + mb + mc ≥ wa f (b, c) + wb f (c, a) + wc f (a, b), 32

where ma , wa are the median, respectively angle bisector corresponding to the side a, etc. (see [1]). We present here a partial solution. We will show that any function f with the property f (x, y) ≤

(x + y)2 4xy

(1)

is a solution. The indicated function s f (x, y) =

x2 + y 2 2xy

clearly satisfies relation (1). For this purpose we prove the following: Lemma. One has in any triangle (b + c)2 ma ≥ . wa 4bc

(2)

Proof. This can be found in our book (see [2], p.112), but we will present it for the sake of completeness. It is well-known that ma wa ≥ s(s − a) (where s is the semi-perimeter). Now, ma ma wa s(s − a) (b + c)2 (b + c)2 = ≥ · = , wa wa2 4 bcs(s − a) 4bc where we used the formula for wa . Therefore, (2) follows. Now, by (2), (b + c)2 ma ≥ wa ≥ wa f (b, c), 4bc so by addition the required property follows.

References [1] M. Bencze, D.M. B˘atinet¸u-Giurgiu, OQ.1338, Octogon Math. Mag., 12(2004), no.1. [2] J. Sándor, Geometric inequalities (Hungarian), Ed. Dacia, Cluj, 1988. 33

9

On a generalization of the Erd¨ os-Mordell inequality

Let M be a point in the interior of a triangle ABC. Let pa , pb , pc denote the distances of M to the sides BC, AC, AB. By searching for functions f such that X X f (M A) ≥ 2 f (pa ) (1) we will obtain a generalization of the famous Erdös-Mordell inequality for a triangle X X MA ≥ 2 pa . (2) It is well-known that MA ≥

c b pb + pc , a a

MB ≥

a c pc + pa , b b

b a CM ≥ pa + pb c c

(see [1], [2]), so if f is an increasing function, we will obtain ¶ X X µc b pb + pc . f (M A) ≥ f a a Now, suppose that the increasing function of positive values f satisfies the following property: f (xy + zt) ≥ xf (y) + zf (t),

x, y, z, t > 0.

(3)

Then, clearly, µ f

c b pb + pc a a

¶ ≥

c b f (pb ) + f (pc ), a a

and by addition we will obtain relation (1). Therefore: Theorem. If the increasing function f satisfies relation (3), then relation (1) holds true. For example, let f (x) = xk (k ≥ 1 real number). Then (xy + zt)k ≥ (xy)k + (zt)k ≥ xy k + ztk Thus we have proved: Corollary. If k ≥ 1 is a fixed real number, then X X M Ak ≥ 2 pka . For k = 1, this is the Erdös-Mordell inequality. 34

for all k ≥ 1.

(4)

References [1] N.D. Kazarinoff, Geometric inequalities, Math. Assoc. America, 1961. [2] J. Sándor, Geometric inequalities, Ed. Dacia, Cluj, 1988.

10

On certain inequalities in a tetrahedron

1. Let ABCD be a tetrahedron in the space, and let AA0 = mA be the median corresponding to the face BCD. Let BA1 and AA1 be two medians of the faces BCD and ACD, respectively.

A

B0 D

B A0

A1

C

Then the median BB 0 (B 0 ∈ ACD) has an endpoint B 0 on the median line AA1 . Since 1 1 A1 B 0 = AA1 , A1 A0 = BA1 , 3 3 clearly A0 B 0 kAB. In the trapezium AB 0 A0 B it is well-known (see e.g. [2]) that if AB 0 ≤ BA0 (i.e. in fact AA1 ≤ BA1 ), then BB 0 ≥ AA0 . Therefore, if AA1 ≤ BA1 , then mA ≤ mB . (1) Now, by using the well-known formula AA1 =

1p 2(AC 2 + AD2 ) − CD2 2 35

etc.

clearly AA1 ≤ BB1 will be equivalent to AC 2 + AD2 ≤ BC 2 + BD2 .

(2)

On other words, if (2) is valid, then (1) is valid, too, i.e. mA ≤ mB . But (2) is in fact a characterization for mA ≤ mB . Indeed, it is well-known (and follows from the above figure) that 2

9AA0 = 3(AB 2 + AC 2 + AD2 ) − (BC 2 + CD2 + BD2 )

(3)

2

9BB 0 = 3(BA2 + BC 2 + BD2 ) − (AC 2 + CD2 + AD2 ) AA0 ≤ BB 0 ⇔ AC 2 + AD2 ≤ BC 2 + BD2 . We will prove that relation (2) holds true in certain particular cases, but not generally. Let us suppose that the height AH of the tetrahedron is in the interior of the tetrahedron, and that AB ⊥ CD. Let AK ⊥ CD. Then clearly BK ⊥ CD, and H ∈ BK.

A

B

D H

K

C

Now by AC 2 = AK 2 + CK 2 ,

AD2 = AK 2 + KD2 ,

BC 2 = BK 2 + CK 2 ,

BD2 = BK 2 + KD2

it follows that AC 2 + AD2 ≤ BC 2 + BD2 ⇔ AK 2 ≤ BK 2 , 36

i.e. AK ≤ BK. Now since SA = area(BCD) = BK · CD/2,

SB = area(ACD) = AK · CD/2,

clearly AK ≤ BK ⇔ SB ≤ SA . Therefore, we have proved that in such tetrahedron one has: mA ≤ mB ⇔ SB ≤ SA .

(4)

By repeating the argument to other faces, relation (4) and its analogous relations give the following result: If ABCD is orthocentric, having acuteangled faces, then mA ≤ min{mB , mC , mD } ⇔ SA ≥ max{SB , SC , SD }.

(5)

This is a strengthening of a particular case of OQ.1324 by M. Olteanu [1]. However, (5) is not true in all tetrahedrons! It is sufficient to show that (2) is not generally true (with SB ≤ SA ). In fact this can be rewritten as AC 2 + CD2 + AD2 ≤ BC 2 + CD2 + BD2 ⇔ SB ≤ SA . Now, in a general triangle having side lengths a, b, c it is well known that X X 16S 2 = 2 a2 b2 − a4 . Since

and denoting

³X X

a2

´2

=

X

a4 + 2

X

a2 b2 ,

a2 = k we get

16S 2 + k 2 = 4

X

a2 b2 = 4c2 (a2 + b2 ) + 4a2 b2 = 4c2 k + 4a2 b2 − 4c4 .

Thus 16S 2 = −k 2 + 4c2 k + 4a2 b2 − 4c2 := f (k).

(6)

The function f (k) (k > 0) has a maximum attained at kmax = 2c2 . Clearly, if 0 < k < kmax , then f (k) is strictly increasing, while for k ≥ kmax , f (k) is strictly decreasing. So, for 0 < k1 , k2 ≤ 2c2 , k1 ≤ k2 ⇒ f (k1 ) ≤ f (k2 ). Remark that k ≤ 2c2 means that a2 + b2 + c2 ≤ 2c2 , i.e. a2 + b2 ≤ c2 (i.e. the b in the triangle is obtuse). Thus when the triangle ACD, BCD are angle C not acute-angled (in A and B), the relation doesn’t hold. 37

A S M R

u

B

D

T

C

2. Now, let us suppose that in the tetrahedron ABCD the faces are acute angled and that BC ⊥ AD. Let AT ⊥ BC. Then clearly T D ⊥ BC and T S ⊥ AD, then we get BS ⊥ AD, CS ⊥ AD. [ \ Let RT A ≡ RT D, i.e. let us consider the bisector plane of the faces ABC and BCD. Similarly, let AM ≡ M D, i.e. one considers the ”median plane” BM C. [ Let m(RT A) = u. Since BS ⊥ AD, BC ⊥ AD, it follows also that AD ⊥ T S. But BC ⊥ AT, AD, so BC ⊥ T S, i.e. T S is the common perpendicular of BC and AD. It follows that ST, T M, T R ⊥ BC, and since T S, T R, T M are all in the plane AT D, then T S, T R, T M are respectively the height, bisector and median of the triangle AT D, corresponding to the vertex T . Therefore T S ≤ T R ≤ T M . Let hBC = area(BSC),

lBC = area(BRC),

mBC = area(BM C).

Then hBC ≤ lBC ≤ mBC which is analogous to the geometry of triangle. Now let V denote a volume. It is well-known (see e.g. [2]) that 2 sin u V (ABCD) = SA · SD · 3 BC 38

(7)

(where SA = area(BCD)). By decomposing the tretrahedron in two smaller tetrahedrons, it follows that u u u u 2 SD · lBC · sin 2 2 SA · lBC · sin 2 2 SA · SD · 2 sin 2 cos 2 · + · = · 3 BC 3 BC 3 BC one gets lBC =

2SA · SD u · cos . SA + SD 2

(8)

But V =

2 sin u hBC · AD = SA · SD · , 3 3 BC

so hBC =

2 SA · SD · sin u · . 3 BC

(9)

Since lBC ≥ hBC , (8) and (9) imply the inequality SA + SD ≤

BC · AD Σ u = u 2 sin sin 2 2

(10)

where Σ is the area of a triangle with base BC and height AD. Note that relation (10) is the extension to the space of the triangle inequality b+c≤

a sin

A 2

(11)

see ([3], Lemma 1). For other inequalities in a tetrahedron, see [2].

References [1] M. Olteanu, OQ.1324, Octogon Math. Mag., 11(2003), no.2, 859. [2] J. Sándor, Geometric inequalities (Hungarian), Ed. Dacia, Cluj, 1988. [3] J. Sándor, On Emmerich’s inequality, Octogon Math. Mag., 9(2001), no.2, 861-864. 39

11

Certain trigonometric inequalities deduced by convexity methods

Trigonometric inequalities are important in many fields of Mathematics (Elementary geometry, Mathematical analysis, Fourier series, Numerical ³ π ánalysis, etc.). While the inequalities like sin x < x, tg x > x, x ∈ 0, are 2 2 well-known, many similar inequalities, like Jordan’s inequality sin x > x are π rediscovered from time to time. For example, in a recent note [1] this, and the x 2 inequality tg < x are proved by using certain auxiliary functions. However, 2 π this way doesn’t suggest a method of discovery of such inequalities. In what follows we shall employ the convexity method to deduce simple (and in many cases important in applications), trigonometric h π i inequalities. 1. Let us consider the concave curve y = sin x on 0, . 2

B       

    

D

A y = sin x

C

√ 2 2 1  2 

      O

π 6

π 4

π 3

π 2

π

2 The equation of line OA is y = x. Since the segment OA is below the π curve, we get: h πi 2 x ≤ sin x, ∀ x ∈ 0, . (1) π 2 π One has equality only for x = 0 or x = . This is the well-known Jordan 2 inequality (see e.g. [2] or [3]). We can improve (1) for smaller intervals. For 40

√ 2 2 example, by observing that the equation of line OB is OB : y = x, one π obtains: √ h πi 2 2 . (2) x ≤ sin x, ∀ x ∈ 0, π 4 √ 2 2 2 Here > , so (2) is indeed an improvement. For the line (BA) the π π angular coefficient is √ 2− 2 √ 2(2 − 2) 2 = , m = tg α = π π 4 therefore

³ π´ y−1=m x− , 2

implying

√ √ √ 2 2( 2 − 1) x + 2 − 1 ≤ sin x. π

Thus:

√ ( 2 − 1)

Ã √ ! hπ π i 2 2 x + 1 ≤ sin x, ∀ x ∈ , π 4 2

(3)

π π with equality only for x = or . We note that this can be written equiva4 2 lently also as √ √ 2 2 ( 2 + 1) sin x ≥ x + 1. (30 ) π Now, by using the line (OC), one can deduce: h πi 3 x, ∀ x ∈ 0, (4) π 6 √ which improves (2) in this interval, since 3 > 2 2. For the equation of line (OD), where √ 3 √ 3 3 2 m= π = 2π 3 one obtains: √ h πi 3 3 sin x ≥ x, ∀ x ∈ 0, . (5) 2π 3 sin x ≥

41

For the line (CB) one has √ 2−1 √ 6( 2 − 1) 2 = m= , π π 12 therefore

√ 1 π´ 6( 2 − 1) ³ y− = x− , 2 π 6

giving:

√ π´ 1 6( 2 − 1) ³ x− + sin x ≥ π 6 2

hπ π i for x in , . This can be transformed, after certain elementary computa6 4 tions, to √ hπ π i √ 6 2−1 , ∀x∈ , ( 2 + 1) sin x ≥ x + . (6) π 2 6 4 ³ π´ 2. Let us consider now the convex function y = tg x, x ∈ 0, . The 2 4 equation of line OA is y = x. π

√

C

3

1 √3

A B

3

π 6

π 4

42

π 3

π 2

This gives, by the convexity of tg x, the following inequality: h πi 4 . (7) tg x ≤ x, ∀ x ∈ 0, π 4 π One has equality only for x = 0 or x = . Let us remark that, by putting 4 h πi and (7) implies 2x = t we get t ∈ 0, 2 h πi t 2 tg ≤ t, ∀ t ∈ 0, . (70 ) 2 π 2 This appears also in [1] (and clearly is equivalent to (7)). By combining (1) and (70 ) we can write: h πi 2 x tg ≤ x ≤ sin x, ∀ x ∈ 0, . (8) 2 π 2 Such inequalities, between different trigonometric functions will appear later, too. Now, for the line (OB) one has √ 3 √ 2 3 3 , m= π = π 6 giving √ h πi 2 3 tg x ≤ x, ∀ x ∈ 0, (9) π 6 √ which strengthens (7) for this interval, since 2 3 < 4. For an extension of (70 ) write the equation √ 3 3 x, (OC) : y = π implying √ h πi 3 3 tg x ≤ x, ∀ x ∈ 0, . (10) π 3 By putting 2x = t, we have: √ · ¸ 3 3 2π t t, ∀ t ∈ 0, (100 ) tg ≤ 2 2π 3 extending inequality (70 ) to a larger interval. For the line (AB) we have √ 3− 3 √ 4(3 − 3) 3 ³ ´ , m= π π = π − 4 6 43

so

4(3 − y−1= π

√ ³ 3) π´ x− , 4

yielding

√ hπ π i √ 4(3 − 3) tg x ≤ x + 3 − 2, ∀ x ∈ , . π 4 6 This can be written also in the form: hπ π i √ √ 24 (3 + 3)tg x ≤ x − (3 − 3), ∀ x ∈ , . π 4 6

(11)

(110 )

3. In the preceding two paragraphs we have deduced inequalities for the sin and tg functions. We now remark that these can be easily transformed π into inequalities for the cos and ctg functions. Applying e.g. (1) for x = − y 2 ³ h π i´ y ∈ 0, one obtains: 2 h πi 2 cos y ≥ 1 − y, ∀ y ∈ 0, , (12) π 2 known also as Kober’s inequality. h πi hπ π i , , we get: In the same manner from (2), since x ∈ 0, gives y ∈ 4 4 2 √ hπ π i √ 2 2 cos y ≥ 2 − y, ∀ y ∈ , . (13) π 4 2 Analogously, from (4) we get cos y ≥

hπ π i 3 3 − y, ∀ y ∈ , 2 π 3 2

(14)

while from (30 ) we can obtain: √ h πi √ √ 2 2 ( 2 + 1) cos y ≥ 2 + 1 − y, ∀ y ∈ 0, π 2

(15)

or in an equivalent form h πi √ 2 (2 − 2)y, ∀ y ∈ 0, . (150 ) π 4 √ We note that this improves (12) since 2 − 2 < 1. Now, inequality (7) π applied to x = − y yields 2 hπ π i 4 ctg y ≤ 2 − y, ∀ y ∈ , . (16) π 4 2 cos y ≥ 1 −

44

In the same manner, from (10) one can prove √ √ hπ π i 3 3 3 3 ctg y ≤ − y, ∀ y ∈ , . 2 π 6 2

(17)

4. We now use tangent lines to concave or convex curves.

(0, 1)

π ,0 4

π ,0 2

³π ´ Writing the tangent equation to the curve y = cos x at the point ,0 2 ³ ´ π since (cos x)0 = − sin x, we have y − 0 = −1 x − . Since the curve is below 2 the tangent, we get h πi π cos x ≤ − x, ∀ x ∈ 0, (18) 2 2 π with equality only for x = 0 or x = . Before going further, we give an 2 application of Jordan’s inequality, combined with (18). By 2 x sin x 4 x tg x = = · > ππ , cos x π π − 2x −x 2 we get Steˇckin’s inequality (see [2]) tg x >

³ π´ 4 x · , ∀ x ∈ 0, . π π − 2x 2 45

(19)

another proof, see [3]. By writing the tangent line equation at the point Ã For √ ! π 2 , to the graph of y = cos x, by the same method we can deduce: 4 2 √ √ h πi 2 π−2 2 . − x, ∀ x ∈ 0, cos x ≤ 4 2 2 Remark that (150 ) and (20) written in a single line, give: √ √ h πi √ 2 π−2 2 2 1 − (2 − 2)x ≤ cos x ≤ − x, x ∈ 0, . π 4 2 4 Ã √ ! π 3 By writing the tangent lines to the curve y = sin x at D , , 3 2

(20)

(21)

√ 3 1³ π´ y= + x− , 2 2 3 we get:

√ x 3 π sin x ≤ + − , ∀ x ∈ [0, π]. 2 2 6

(22)

³π ´ , 1 since By writing the tangent line to the convex curve y = tg x at A 4 ³ ´ 1 π (tg x)0 = , we get y − 1 = 2 x − , implying 2 cos x 4 ³ π´ π tg x ≥ 2x − + 1, ∀ x ∈ 0, (23) 2 2 π . We note here that (23) and (19) cannot be 4 compared (i.e. one of them doesn’t imply the other one for all x). 5. The above methods can be applied to inverse trigonometrical functions, too. Considering e.g. the concave curve y = arctg x on [0, ∞) one obtains easily with equality only for x =

arctg x ≥

π x, ∀ x ∈ [0, 1]. 4

(24)

³ π´ By writing the tangent line equation in the point 1, , by (arctg x)0 = 4 1 π 1 , one obtains y − = (x − 1), giving 2 1+x 4 2 arctg x ≤

x−1 π + , ∀ x ∈ [0, ∞). 2 4 46

(25)

For the³convex curve y = arcsin x, x ∈ [0, 1] by writing the line passing on π´ (0, 0) and 1, , we get 2 π (26) arcsin x ≤ x, ∀ x ∈ [0, 1] 2 Ã√ ! 2 π while by writing the tangent line equation at , we get 2 4 arcsin x ≥

√ π 2x + − 1, ∀ x ∈ [0, 1]. 4

(27)

More generally (and this can be done in all cases) for an arbitrary x0 ∈ (0, 1) one can obtain x − x0 arcsin x ≥ arcsin x0 + p , ∀ x ∈ [0, 1]. 1 − x20

(28)

6. Another argument is to approximate sin or cos by a quadratic function. 1 Let us consider f (x) = ax(π − x). Since f 0 (0) = πa, we will select a = . π Then f 0 (π) = −1 as for the sin function. Now it is immediate that the graph of f is below the graph of sin on [0, π], giving: sin x ≥

x(π − x) , ∀ x ∈ [0, π]. π

(29)

Such simple inequalities have importance in Fourier series. Remark that x(π − x) 2x π ≥ ⇔ π − x ≥ 2 i.e. x ≤ π − 2 (π − 2 < by π < 4). Thus for π π 2 x ∈ [0, π − 2], relation (29) is better than Jordan’s inequality (1). Applying the same procedure to the cos function and ³ π´ π´³ x+ , a > 0, f (x) = −a x − 2 2 1 since (cos x)0x= π = −1, we select a = , and as above one can deduce the 2 π inequality h π πi 1 π cos x ≥ − x2 + , ∀ x ∈ − , (30) π 4 2 2 The functions sin and cos can be approximated by more complicated functions, too. For example, by taking µ 2 ¶ π − x2 f (x) = x , π 2 + x2 47

one can deduce Redheffer’s inequality (see [2] or [3]) sin x π 2 − x2 , ∀ x ∈ [0, π]. ≥ 2 x π + x2 By considering

µ f (x) = x

inequalities of type

µ

sin x x

¶a ≤

π a − xa π a + xa

(31)

¶1

a

,

h πi π a − xa , ∀ x ∈ 0, π a + xa 2

(32)

can be proved, see [4]. 7. Finally, we apply the well-known Jensen-Hadamard (or HermiteHadamard, or Hadamard) inequalities for concave (or convex) functions f : [a, b] → R (see e.g. [5]): µ · ¸ ¶ Z a+b f (a) + f (b) ≤ b ≤ f (x)dx( 0. For n = 7 and 49

π − t where t > 0 and t → 0. Then (1) reads 2 ³π ´ ³n ³π ´´ sin −t 2 cos −t ≤ . π 2 2 −t 2 √ 2 2 Now t = 0 yields for n = 7 and n = 8 the two invalid inequalities ≤ 2 π 2 and 1 ≤ , resp. Therefore a continuity-argument shows that (1) cannot be π true in general. ³x´ sin x x x (ii) For n = 1 inequality (1) becomes cos ≤ , that is ≤ sin . 2 x 2 2 This shows that (1) is false for n = 1, too. (iii) On the other hand, for n = 2, 3, 4, 5, 6 inequality (1) is true. In order to prove this assertion we employ a special case of a general majorization inequality, namely: If we suppose that m ≤ xi ≤ M , i = 1, . . . , n then there exists a unique w ∈ [m, M ) and a unique integer N ∈ {0, 1, . . . , n} such that: n = 8, we let x1 = · · · = xn =

n X

xi = (n − N − 1)m + w + N M.

i=1

Then for any convex function f : [m, M ] → R it holds n X

f (xi ) ≤ (n − N − 1)f (m) + f (w) + N f (M )

(2)

i=1

(see [2], p.64 and 132). Now, it is well-known that f (x) =

x , sin x

0≤x≤

π , 2

is convex. Now the proof of the five mentioned cases ´ of n runs as follows: ³ π π yields for x1 + · · · + x5 = N + w, Let e.g. n = 5. Then x1 , . . . , x5 ∈ 0, 2 2 π 0 ≤ w < the five subcases N = 0, 1, 2, 3, 4 with the corresponding five 2 inequalities µ ¶ Nπ w 5 π cos + ≤ , 0≤w< . (3) π w 4 2 2 4−N +N + 2 sin w 50

(In order to estimate the sums

5 X xk we employ inequality (2) and sin xk k=1

x = 1.) Now (3) is immediate for N = 2, 3, 4 as its left-hand side lim x→0 sin x is negative. The cases N = 0 and N = 1 are checked by a computer-algebra system as are the respective inequalities for the remaining cases of n. We summarize the above considerations as: Theorem. Inequality (1) is valid if n ∈ {2, 3, 4, 5, 6}. Remark. Inequalities (3) for N = 0 and general n ∈ N deserve special attention. They read cos

w n sin w ≤ , 2 w + (n − 1) sin w

that is w + (n − 1) sin w ≤ 2n sin

w , 2

(4)

π . 2 We claim that (4) is valid whenever n ≥ 2 even for 0 ≤ w ≤ π. Indeed, if we let w f (w) = 2 sin − (n − 1) sin w − w, 2

where 0 ≤ w ≤

then f 0 (w) = n cos

w w w − (n − 1) cos w − 1 = −2(n − 1) cos2 + n cos + n − 2 2 2 2 ³ ´ w´³ w = 1 − cos 2(n − 1) cos + n − 2 , 2 2

whence f 0 (w) ≥ 0. This and f (0) = 0 yields the claim. Furthermore, as f (w), 3π , firstly increases and then decreases we get due to 0≤w≤ 2 µ ¶ √ 3π 3π f = n( 2 + 1) − −1: 2 2 3π Inequality (4) is valid for 0 ≤ w ≤ wn , where wn > , when = even 2 n ≥ 3. 51

References [1] M. Bencze, OQ.1216-1218, Octogon Math. Mag., 11(2003), no.1, 390. [2] A.W. Marshall, I. Olkin, Inequalities - Theory of Majorization and its applications, Academic Press, New York, 1979. [3] W. Janous, On some trigonometric inequalities of M. Bencze, Octogon Math. Mag., 11(2003), no.2, 720-724.

13

On

sin x x

We will determine all a ∈ R such that ¶ µ ³ π´ sin x a π a − xa ≤ a for all x ∈ 0, . x π + xa 2

(1)

We may assume a > 0 since for a ≤ 0 clearly π a − xa < 0. Then (1) can be written also as µ a ¶1 π − xa a sin x ≤ f (x) = x (2) π a + xa where we extend the definition of f to [0, π]. After certain elementary computations one can deduce that the derivative f is µ 0

f (x) =

π a − xa π a + xa

¶1 µ a

−x2a − 2π a xa + π 2a π 2a − x2a

¶ .

(3)

Since by putting π a =√λ, xa = t, the quadratic equation −t2 − 2λt + t2 = 0 √ has the solutions t1 = λ( 2 − 1), t2 = −λ( 2 + 1), one can deduce √ that the a numerator √ of the second term in (3) can be decomposed as [π ( 2 − 1) − xa ][π a ( 2 + 1) + xa ]. Therefore we have obtained that f (0) = 0, f (π) = 0 √ π and f has a maximum at x0 = √ (= π( 2 − 1)). To prove that (2) ³ π ´2 + 1 π π < holds true we must show that f ≥ 1 (since √ and otherwise 2 2 2+1 ³ π´ there exists x1 ∈ 0, with sin x1 > f (x1 ) - contradiction). The inequality 2 µ a ¶1 ³π ´ 2 −1 a π f ≥ 1 becomes equivalent with ≥ , a > 0. Since the function a 2 2 +1 2 µ a ¶1 a 2 −1 g : (0, ∞) → R, g(a) = is strictly increasing (which can be easily 2a + 1 52

shown with the derivative of log g) and lim g(a) = 1, there exists a single a→∞ 1 2 a0 > 0 such that g(a0 ) = < 1. Here a0 > 1, otherwise g(a0 ) ≤ g(1) = , i.e. π 3 2 1 ≤ (6 ≤ π) - contradiction. Any a ≥ a0 satisfies inequality (1). We note π 3 r 3 that a0 < 2, since for a0 ≥ 2 we would obtain g(a0 ) ≥ , i.e. 16 ≥ 3π 2 4 contradiction. Remark. Since a = 2 is a solution of (1), by Redheffer’s inequality ([1], sin x π 2 − x2 [2]) ≥ 2 we can write the following double inequality: x π + x2 µ

sin x x

¶2 ≤

π 2 − x2 sin x , ≤ 2 2 π +x x

³ π´ x ∈ 0, . 2

(4)

References [1] D.S. Mitronović, Analytic inequalities, Springer Verlag, 1970. [2] J. Sándor, Geometric inequalities (Hungarian), Ed. Dacia, Cluj, 1988.

14

On de Cusa’s trigonometric inequality

According to F.T. Câmpan ([1]) the following trigonometric inequality was discovered by Nicolaus de Cusa ³ (1401-1464): π´ Theorem 1. For all x ∈ 0, one has 2 3 sin x < x. 2 + cos x Such inequalities (along with e.g. 2 sin x + tg x > 3x, known as Huygens’ inequality, see e.g. [1], [2]) were studied by geometric arguments also by Willebrod Snellius (1581-1626), and Christian Huygens (1629-1695). We shall extend de Cusa’s inequality as follows: Theorem 2. Let a, b, c > 0 such that 3b ≤ c ≤ a + b. Then for any ³ π´ x ∈ 0, one has 2 c sin x < x. a + b cos x

53

h π´ Proof. Let us consider the application f : 0, → R, 2 f (x) = x(a + b cos x) − c sin x. One can remark that f (0) = 0, f 0 (x) = a + b cos x − bx sin x − c cos x,

f 0 (0) = a + b − c ≥ 0

by assumption. Moreover, f 00 (x) = (sin x)(c − 2b) − bx cos x = (cos x)[(c − 2b)tg x − bx]. h π´ Put g(x) = (c − 2b)tg x − bx, x ∈ 0, . By g(0) = 0, g 0 (x) = 2 (c − 2b) − b cos2 x , since b cos2 x − (c − 2b) ≤ b − (c − 2b) = 3b − c ≤ 0 we get 2x cos g 0 (x) ≥ 0, i.e. g is an increasing function. Thus g(x) > g(0) for x > 0. This in turn implies f 00 (x) > 0, so f 0 (x) > f 0 (0) ≥ 0 for x > 0, i.e. f 0 (x) > 0. Thus f (x) > f (0) = 0 for x > 0, and this finishes the proof of the theorem. Remark. For a = 2, b = 1, c = 3 we reobtain Theorem 1.

References [1] F.T. Câmpan, The story (fairy tale) of the number π (Romanian), Ed. Albatros, Romania, 1977. [2] J. Sándor, Geometric inequalities (Hungarian), Ed. Dacia, Cluj, 1988.

15

A property of sin

1 x

We will determine a class of functions f : R∗+ → R∗+ so that f (x) → 0 (x → ∞) and x2 [f (x) − f (x + 1)] → 1 (x → ∞). An example is given by 1 f (x) = sin . x Theorem. If f is differentiable and x2 f 0 (x) → l (x → ∞), then x2 [f (x) − f (x + 1)] → −l. Proof. By the Lagrange mean-value theorem we can write f (x) − f (x + 1) = [x − (x + 1)]f 0 (ξ), where x < ξ < x + 1. By Now,

x x 1 x < 1 < + , we get ξ → ∞ and → 1 as x → ∞. ξ ξ ξ ξ

µ ¶2 x x [f (x) − f (x + 1)] = −x f (ξ) = − (ξ 2 f 0 (ξ)) → −1 · l = −l, ξ 2

2 0

54

by the given assumption. This proves the theorem. 1 1 Remark 1. For f (x) = sin one has x2 f 0 (x) = − cos → −1 as x → ∞, x x so x2 [f (x) − f (x + 1)] → 1. Remark 2. The above result (in Remark 1) could be deduced also by the L’Hôpital rule.

16

(sin x)cos x + (cos x)sin x made integer Let A = (sin x)cos x + (cos x)sin x ∈ Z. First note that we must suppose sin x > 0, cos x > 0. Then clearly 0 < (sin x)cos x ≤ 1,

0 < (cos x)sin x ≤ 1.

Thus 0 < A ≤ 2. On the other hand, A has the form A = ab + ba , where a, b > 0. It is well-known that ab + ba > 1 for all a, b > 0 (see e.g. [1], p.281). Since A is integer, this implies A ≥ 2. Therefore, we must have A = 2, i.e. (sin x)cos x = 1,

(cos x)sin x = 1.

This implies sin x = cos x = 1, which is impossible. Therefore A 6∈ Z, for all x ∈ R.

References [1] D.S. Mitrinović, Analytic inequalities, Springer Verlag, 1970.

55

56

Chapter 2

Sequences and series of real numbers ”... The infinite we shall do right away. The finite may take a little longer.” (Stanislaw M. Ulam)

”... Mathematics is concerned only with the enumeration and comparison of relations.” (Carl F. Gauss)

57

1

On certain limits related to the number e

1. In the very interesting web pages by Steven Finch ([3]) there appears also the following limit, submitted by Felix A. Keller: · ¸ nn (n − 1)n−1 lim − = e. (1) n→∞ (n − 1)n−1 (n − 2)n−2 In what follows we will show how certain expressions which are similar to the bracket in (1), are intimately related to certain special means of two arguments, namely the logarithmic and identric means. Let a, b > 0 be two positive real numbers. The logarithmic, resp. identric means of a and b are defined by L = L(a, b) =

b−a ln b − ln a

(a 6= b),

L(a, a) = a

and

1 I = I(a, b) = (bb /aa )1/(b−a) (a/neb), I(a, a) = a. e For many properties, generalizations, etc. of these means we quote e.g. [1], [2], [5], [8], [10]. The following inequalities, which will be used here, are due to K.B. Stolarsky [10]: √ a+b ab < L(a, b) < I(a, b) < (a 6= b). (2) 2 These relations can be much improved, see e.g. [8]. 2. In what follows, we will use the following notations: µ ¶ µ ¶ 1 x 1 x f (x) = 1 + , F (x) = (x + 1) 1 + , x x µ ¶ 1 1 g(x) = ln 1 + − (x > 0). x x+1 Since ln I(x, x + 1) = (x + 1) ln(x + 1) − x ln x − 1 and 1 = ln(x + 1) − ln x, L(x, x + 1) we easily can see that ln(F (x)) = ln(x + 1) + ln f (x) = ln eI(x, x + 1), 58

(3)

so F (x) = eI(x, x + 1) = (x + 1)f (x) and F 0 (x) =

F (x) L(x, x + 1)

(4) (5)

where F 0 denotes the derivative of F . 3. We note that by (3) we have f (x) =

e I(x, x + 1), x

so by inequalities (2) applied to a := x, b := x + 1, we get the following double-inequality r x 2x + 1 e < f (x) < e . (6) x+1 2x + 2 As a consequence, we can write x x √ e < x[e − f (x)] < e, √ 2x + 2 x+ x+1

(7)

yielding the following limit: e lim x[e − f (x)] = . x→∞ 2

(8)

Of course, this limit may be obtained by l’Hôpital rule, too. However, the inequalities provide a much better information. For another proof of (7), based on Hadamard’s inequality, see [7]. Remark. By nf (n) − (n − 1)f (n − 1) = n[f (n) − e] + (n − 1)[e − f (n − 1)] + e, (8) implies lim [nf (n) − (n − 1)f (n − 1)] = e

n→∞

(80 )

proved by other arguments in [6]. The limit (80 ) is due to M. Gherm˘anescu [4]. 4. We now prove first the following result: Theorem 1. For x > 0, F is a strictly increasing, strictly concave function. Proof. Clearly, F (x) F 0 (x) = > 0, L(x, x + 1) 59

and, on the other hand since · F (x) = F (x) (ln(x + 1) − ln x)2 − 00

1 x(x + 1)

¸

·

¸ 1 1 = F (x) 2 − , L (x, x + 1) x(x + 1) by the left side of (2) we clearly get F 00 (x) < 0. Thus F is strictly concave. Corollary 1. For all positive integers n > 1 one has r 2n n f (n) < F (n + 1) − F (n) < f (n − 1) . (9) 2n + 1 n+1 Proof. By Lagrange’s mean value theorem one has F (n + 1) − F (n) = F 0 (ξ)

(ξ ∈ (n, n + 1)),

and by Theorem 1, F 0 is strictly decreasing, so we have F (n) F (n + 1) < F (n + 1) − F (n) < . L(n + 1, n + 2) L(n, n + 1)

(10)

Now, (9) follows by (10) and (2), by remarking that F (n) = (n + 1)f (n) and

2n n < < 2n + 1 L(n, n + 1)

r

n . n+1

Corollary 2. lim [F (n + 1) − F (n)] = e. n→∞

(11)

This follows from (10), or (9), if we remark that f (n) → e (n → ∞). Thus the limit (1) follows. Remark. Clearly, (9)-(10) are valid for all positive real numbers n > 1. 5. The bracket in limit (1) can be written also as nf (n − 1) − (n − 1)f (n − 2) = n[f (n − 1) − f (n − 2)] + f (n − 2). Since f (n − 2) → e (n → ∞), we clearly get (from (1) or (11) that lim n[f (n − 1) − f (n − 2)] = 0.

n→∞

60

(12)

This could be proved also by l’Hôpital rule, but we are interested for more precise relations of type (6), or (9), (10). These in turn allow us to study interesting problems. Theorem 2. The function g is strictly positive, strictly decreasing for x > 0, and all x > 0, the following inequality is valid: (g(x))2 + g 0 (x) < 0. Proof. Since g(x) =

(13)

1 1 − , L(x, x + 1) x + 1

the inequality g(x) > 0 is a simple consequence of the fact that L is a mean (or the right-side of (2)). A simple computation shows that g 0 (x) =

−1 < 0. x(x + 1)2

In order to prove (13), calculate ¸ · 1 1 1 2 0 − − (g(x)) + g (x) = √ x+1 p This is true, since L(x, x + 1) > x(x + 1) (see (2)) it is sufficient to prove that √ p (x + 1) x x(x + 1) > √ , x+1 √ √ or equivalently x + 1 > x + 1. Corollary 3. The function f is strictly increasing and strictly concave for all x > 0. Proof. Since f 0 (x) = f (x)g(x),

f 00 (x) = f (x){[g(x)]2 + g 0 (x)},

this is a simple consequence of Theorem 2 (i.e. (13)). Corollary 4. Let (an ) be a strictly positive sequence such that lim an g(n) = l.

n→∞

61

(14)

Then lim an [f (n) − f (n − 1)] = le.

n→∞

(15)

Proof. By the Lagrange mean-value theorem one has f (n) − f (n − 1) = f 0 (ξ) = f (ξ)g(ξ), and since f is strictly concave, f g is strictly decreasing, yielding the double inequality f (n)g(n) < f (n) − f (n − 1) < f (n − 1)g(n − 1) (n > 1).

(16)

Now (15) follows from (16), (14), and the remark that lim g(n − 1)g(n) = 1

n→∞

(this is implied by (2), but also by simple application of l’Hôpital rule). Examples. 1) Let an = n. Then an g(n) → 0 (more generally, xg(x) → 0 as x → ∞), so from (15) we reobtain (12). 1 2) Let an = n2 . Then, since x2 g(x) → (x → ∞), we obtain 2 e lim n2 [f (n) − f (n − 1)] = . (17) n→∞ 2 Remark. This limit is due to Gh. Stoica [9]. 6. Since f is strictly concave, we have that the sequence (f (n + 1) − f (n)) is strictly decreasing. Similarly, the sequence (F (n + 1) − F (n)) is strictly decreasing, too. We consider the sequence given by the limit (8). Theorem 3. Let h(x) = f (x)[1 + xg(x)] (x > 0). Then the function h is strictly increasing. Proof. Since f 0 (x) = f (x)g(x), after a simple calculus we get h0 (x) = f (x)[2g(x) + x(g 2 (x) + g 0 (x))].

(18)

1 1 − (where L = L(x, x + 1) for simplified notation), L x+1 the square bracket in (18) is Since g(x) =

P (y) = xy 2 + 2y − where y=

1 , (x + 1)2

1 1 − . L x+1 62

Since the roots of the polynomial P (y) are y1,2

p −(x + 1) ± x + (x + 1)2 = x(x + 1)

it is sufficient to prove that 1 1 − > L x+1

p

or L f (n − 1) + f (n + 1). Proof. Since u is strictly concave, we have u(n + 1) − u(n) < u(n) − u(n − 1), which transforms into a). On the other hand, by Theorem 1, F (x) = (x + 1)f (x) is strictly concave, too, thus F (n + 1) − F (n) < F (n) − F (n − 1), giving b). Finally, relation c) is a consequence of Corollary 2 (i.e., the function f is strictly concave). 63

References [1] H. Alzer, Ungleichungen f¨ ur Mittelwerte, Arch. Math., 47(1986), 422426. [2] B.C. Carlson, The logarithmic mean, Amer. Math. Monthly, 79 (1972), 615-618. [3] S. Finch, Favorite mathematical constants. The number e, See http://www/mathsoft.com/asolve/constant/constant.html. [4] M. Gherm˘anescu, Problem 4600, Gaz. Mat., 41(1935), 216. [5] A.O. Pittenger, Inequalities between arithmetic and logarithmic means, Univ. Beograd Publ. Elektr. Fak. Ser. Mat. Fiz., 680(1980), 15-18. [6] T. Popoviciu, On the calculation of certain limits (Romanian), Gaz. Mat. Ser. A, vol.LXXVI, no.1, 1971, 8-11. [7] J. Sándor, On Hadamard’s inequality (Hungarian), Mat. Lapok, 87 (1982), 427-430. [8] J. Sándor, On the identric and logarithmic means, Aequationes Math., 40(1990), 261-270. [9] Gh. Stoica, Problem C:325, Gaz. Mat. 8/1983, 358. [10] K.B. Stolarsky, The power and generalized logarithmic means, Amer. Math. Monthly, 87(1980), 545-548. Note added in proof. After completing this paper we have learned that Professors H. Brothers and J.A. Knox have proved relation (1) by using certain series expansions for e. The author is grateful to the authors for calling his attention to the following papers: 1) H.J. Brothers and J.A. Knox, New closed form approximations to the logarithmic constant e, Mathematical Intelligencer 20(1998), no.4, 25-29; 2) H.J. Brothers and J.A. Knox, Novel series-based approximations to e, College Math. Journal, vol.30, 1999, 269-275.

2

On some strange sequences

A. Let (un ) be a sequence defined by µ ¶ 1 n+un 1+ = e, n 64

n = 1, 2, . . .

(1)

Then un =

Put

1 µ ¶ − n. 1 ln 1 + n

1 = t, and consider n f (t) =

1 1 t − ln(1 + t) − = . ln(1 + t) t t ln(1 + t)

Since ln(1 + t) = t −

t2 t3 + − ..., 2 3

one has 1 t t2 t2 t3 t4 − + − ... − + − ... 3 4 4 = 2 3 = a0 + a1 t + a2 t2 + . . . f (t) = 2 4 2 3 t t t 1 t2 − + − . . . 1 − + − ... 2 3 2 3 Then µ ¶ t t2 1 t t2 1 − + − . . . (a0 + a1 t + a2 t2 + . . . ) = − + − . . . 2 3 2 3 4 ∞ X

Let the power series (obtained by Cauchy-product) on the left side be cn tn . Then clearly

n=1

1 1 a1 (−1)n a0 1 cn = an − an−1 + an−2 − · · · + (−1)n−1 + = (−1)n (2) 2 3 n n+1 n+2 1 1 by identification. Since a0 = , c0 = , we get successively 2 2 1 1 c1 = a1 − a0 = − , 2 3 so a1 =

1 1 1 − =− , 4 3 12

so a2 =

1 1 1 c2 = a2 − a1 + a0 = , 2 3 4

1 1 1 1 − − = ,... 4 24 6 24 65

We have obtained un =

1 1 1 + a1 + a2 2 + . . . , 2 n n

(3)

1 (n → ∞) 2 and (3) gives the asymptotical development. The sequence (xn ) defined by x2n+1 = un , x2n = un − 1 in OQ.805 therefore cannot be convergent. Application. We will determine a, b ∈ R, so that µ ¶ 1 a n un − → b as n → ∞ (see [1]) (1) 2 where all (ai ) are recurrently determined by (2). Thus un →

In a recent note [2] we have determined the asymptotic expansion of the sequence (un ), given by un =

1 a1 a2 + + 2 + ... 2 n n

(2)

where a1 , a2 , . . . are constants (which can be effectively computed). Now (2) implies ¶ µ 1 1 → a1 = − . (3) n un − 2 12 Therefore, as an answer to the problem (1), the following can be stated: 1 if a = 1, then b = a1 . If a > 1, then b = −∞. Since un − → 0, clearly for 2 a ≤ 0 one has b = 0. For a ∈ (0, 1) from (3) it follows that b = 0.

References [1] M. Bencze, D. M. B˘atinet¸u-Giurgiu, OQ.1023, Octogon Math. Mag., 10(2002), no.2, 1052. [2] J. Sándor, On the Open Problem OQ.805, Octogon Math. Mag., 10 (2002), no.1, 416.

B. Let (xn ) be a sequence defined by · µ ¶ ¸n+xn 1 1 n 1 1+ =√ e n e 66

(1)

By taking logarithms, one obtains · µ ¶¸ 1 1 (n + xn ) −1 + n ln 1 + =− , n 2 thus xn =

Put t =

1 µ ¶¸ − n. 1 2 1 − n ln 1 + n ·

(2)

1 , and consider the expression n 2 t − 2 + ln(1 + t) 1 t ¸− = · ¸ = A. · 1 1 t 2 1 − ln(1 + t) 2t 1 − ln(1 + t) t t 1

Since ln(1 + t) = t −

t2 t3 + − ... 2 3

we get µ ¶ 2 t2 t3 2 2 t3 t−2+ t − + − ... t − 2 + 2 − t + t − + ... t 2 3 3 2 µ ¶¸ A= · = 4 1 t2 t3 2 t 3 2t 1 − t − + − ... 2t − 2t + t2 − t + − . . . t 2 3 3 2 2 2 t3 t − + ... 2 3 2 = = + a1 t + a2 t2 + . . . , 4 3 2 t t2 − t3 + − . . . 3 2 where the coefficient ai (i = 1, 2, . . . ) can be determined from the equality µ ¶µ ¶ 2 3 t4 2 2 2 t3 2 2 t − + ··· = t − t + − ... + a1 t + a2 t + . . . . (3) 3 2 3 2 3 For this method, based on Cauchy-product of series, see our note [2]. Thus a1 , a2 , . . . , can be effectively computed (e.g. a1 = −1/18). All in all, the asymptotical expansion of xn is xn =

2 1 1 + a1 + a2 2 + . . . 3 n n

2 Particularly, limn→∞ xn = , answering OQ.863 of [1]. 3 67

(4)

References [1] M. Bencze, OQ.863, Octogon Math. Mag., 10(2002), no.1, 518. [2] J. Sándor, On the Open Problem OQ.805, Octogon Math. Mag., 10 (2002), no.1, 416.

C. Let (xn ) be determined by Ãµ n

1 1+ n

¶−n+xn

1 − e

! =

1 . 2e

(1)

An easy computation gives µ ¶ 1 ln 1 + −1 2n ¶ + n. µ xn = 1 ln 1 + n Put

1 = x, and consider the expression n ³ ³ x´ x´ ln 1 + −1 1 x ln 1 + −x+1 2 2 A= + = ln(1 + x) x x ln(1 + x)

x2 x3 x4 x5 x2 x3 x4 − + − + · · · − x + x − + − + ... 2 · 22 3 · 23 µ 4 · 24 2 3 4 ¶ = 2 2 3 x x x x2 1 − + − + ... 2 3 4 µ ¶ 5 5 59 2 5 3 5 4 59 5 x − x+ x − ... x − x + x − ... 24 24 320 24 320 ¶ = 24µ = =B x x2 x3 x x2 x3 2 x 1− + − + ... 1− + − + ... 2 3 4 2 3 4 where we have used the following well-known formulae: ln(1 + x) = x −

x2 x3 + − ... 2 3

³ x´ x x2 x3 ln 1 + = − + − ... 2 2 2 · 22 3 · 23 68

(2)

for x in a neighborhood of 0. Now B can be written as B = x(a0 + a1 x + a2 x2 + . . . ), where a0 , a1 , a2 , . . . can be determined by considering the Cauchy product of power series (see [2]): µ ¶ x x2 x3 5 5 59 2 2 (a0 + a1 x + a2 x + . . . ) 1 − + − + ... = − x+ x −... 2 3 4 24 24 320 For example, a0 =

5 5 , a1 = − , . . . Therefore, 24 48 ³ ´ 1 a1 a2 xn = a0 + + 2 + ... , n n n

(3)

implying e.g. that xn → 0 nxn → a0 =

(4)

5 24

etc.

(5)

References [1] M. Bencze, OQ.860, Octogon Math. Mag., 10(2002), no.1, 517. [2] J. Sándor, On the Open Problem OQ.805, Octogon Math. Mag., 10 (2002), no.1, 416.

D. Let (xn ) be a sequence defined by the relation: µ ¶ 1 1 1 2n + 1 + + ··· + = logxn . n n+1 2n n Since

µ

µ logxn clearly

(1)

2n + 1 n

¶

¶ 2n + 1 ln n = , ln xn

¶ µ 1 ln 2 + n , ln xn = c2n − cn 69

(2)

where

1 1 + ··· + . (3) 2 n Since cn = ln n + γ + o(1), it is well-known that c2n − cn = ln 2 + o(1). Then (2) implies xn → e (n → ∞) (4) cn = 1 +

Now, by Taylor expansion it is immediate that µ ¶ 1 1 1 1 1 1 1 ln 2 + = ln 2 + · − 2· + 3· − ... 2 n n 1·2 n 2·2 n 3 · 23

(5)

On the other hand, it is well-known that (and this can be deduced e.g. by the Euler-Maclaurin summation formula) cn = 1 +

1 1 1 1 1 + · · · + = γ + ln n + + α1 2 + α2 4 + . . . 2 n 2n n n

(6)

B2 B1 , . . . with (Bi ) the Bernoulli numbers), we (where in fact α1 = − , α2 = 2 4 get c2n − cn =

1 1 1 1 +···+ = ln 2 + b1 + b2 2 + . . . n+1 2n n n

1 (b1 = − , etc.). (7) 4

Now, by (2) ln xn =

ln 2 + a1 x + a2 x2 + . . . = 1 + c1 x + c2 x2 + . . . ln 2 + b1 x + b2 x2 + . . .

(8)

1 ) (where (ai ) are from (5), while (bi ) from (7)). Now (ci ) can be n determined by the Cauchy-product of two series and a simple recurrence (for this method see [1]).

(where x =

References [1] J. Sándor, On the open problem OQ.805, Octogon Math. Mag., 10(2002), no.1, 416.

E. Let (xn ) be determined by the relation µ ¶n+xn n = n!, en 70

(1)

µ ¶ 1 n where en = 1 + . n By logarithmation we get xn =

ln n! µ ¶ − n. 1 ln n − n ln 1 + n

(2)

Since ln n! = and

µ ¶ 1 a1 a2 n+ ln n − n + a0 + + 2 + ... 2 n n

(Stirling series)

¶ µ 1 1 1 1 = − 2 + 3 − ... ln 1 + n n 2n 3n

from (2) we can write, after a short computation: 1 c1 c2 ln n + c0 + + 2 + ... n n xn = 2 a1 a2 ln n + a0 + + 2 + ... n n

(3)

√ 1 Here all coefficients are known (e.g. a0 = 1, c0 = ln 2π − , etc.). There2 fore, from (3) xn can be computed with as higher accuracy as we wish. For example, (3) implies µ ¶ A 1 1 +O (4) xn = + 2 ln n n ln n µ ¶ 1 1 implying also xn → (n → ∞), xn − ln n → A (n → ∞), etc. 2 2

F. Let (xn ) be a sequence given by 1+

1 1 + · · · + − ln(n + xn ) = γ, 2 n

1 , and will find 2 1 the asymptotical evaluation of (xn ). The result with lim xn = as for as n→∞ 2 where γ is Euler’s constant. We will prove that lim xn = n→∞

71

we know, was proposed also as a problem in Gazeta Matematic˘a (Bucure¸sti). Here we obtain a new solution. De Tempe [1] gave an elementary proof that: ¶ µ n X 1 1 1 1 < −γ < . (1) − ln n + 2 24(n + 1) k 2 24n2 k=1

This gives the double inequality µ µ ¶ ¶ 1 1 1 1 2 n+ e 24(n+1) − n < xn < n + e 24n2 − n. 2 2

(2)

Now, it is not difficult to prove (e.g. with l’Hôpital rule) that each side 1 of (2) has a limit equal to as n → ∞. This proves that (xn ) is convergent, 2 1 having as a limit . 2 Negoi [2] has applied De Tempe’s method to prove that µ ¶ n X 1 1 1 1 1 − − ln n + + −γ 0

for s > 1 (where Λ is the von Mangoldt function, see [1]) we have A0 (s) > 0, ∀ s > 1, i.e. A is strictly increasing in this domain. Therefore A(n) < π 2 , implying ¯ ¯ ¯ B4n · B4n−2 ¯ ¯ ¯ < π 2 xn . ¯ ¯ 4 B2n

References [1] T.M. Apostol, Introduction to analytic number theory, Springer Verlag, 1976. [2] J. Sándor, On Stirling’s formula (Romanian), Lucr. Semin. Did. Mat., 14(1998), 235-239.

8

The number of factorizations of n

Let f (n) be the number of ”essentially different factorizations” of n, i.e. the number of representations of the positive integer n as a product of integral factors larger than 1, a change in the order of factors not counting as a distinct representation. In OQ.528, proposed by M. Bencze it is conjectured that f (n) ≤ n −

1 for n ≥ 10. n!

In fact, much stronger relations are known in the literature. W. Chen [2] proved that n f (n) ≤ + 1 for all n. 4 113

Clearly

n 1 3n 1 +11+ . 4 n! 4 n! 1 1 1 This is true for all n ≥ 3. (Indeed, ≤ n ≤ for n ≥ 4, while n! 2 2n 3n 1 2n + 1 >1+ = since 3n2 > 4n + 2. For n = 3, a direct verification is 4 2n 2n valid). F.W. Dodd and L.E. Matties [3] proved that f (n) ≤

n (log n)α

for all fixed α > 0 and n sufficiently large; if the least prime factor of n is > 3, then f (n) < n/ log n, proved by H.Z. Cao [4]. It is conjectured that n for n 6= 144, but this is open, as far as know. For other results, f (n) ≤ log n see [5].

References [1] M. Bencze, OQ.528, Octogon Math. Mag., 8(2002), no.2, 622. [2] W. Chen, Upper bound if the number of multiplicative partitions (Chinese), Acta Math. Sinica, 32(1989), 604-609. [3] F.W. Dodd, L.E. Matties, Estimating the number of multiplicative partitions, Rocky Mountain J. Math., 17(1987), 797-813. [4] H.Z. Cao, On a conjecture of multiplicative partitions, Rend. Mat. Appl. 7, Ser.11, no.4(1991), 729-735. [5] J. Sándor, D.S. Mitrinović, Handbook of number theory, Kluwer Academic Publishers, 1996.

9

On perfect and m-perfect numbers

This note is in connection with the paper [1] and the Open Problem OQ.113 (Octogon Math. Mag., 5(1997), no.2). First we wish to note that the terminology of ”very perfect numbers” is not known, but the ”superperfect numbers” one is well-known. Thus, D. Suryanarayana [7] calls a number n superperfect if σ(σ(n)) = 2n and he and H.J. Kanold [7], [4] obtain the general form of even superperfect numbers. Kanold [4] proves that all odd superperfect numbers are perfect 114

squares, but we do not know if there exists at least one number of this type. D. Suryanarayana [8] shows that there is no superperfect number of the form p2α , and this is more general than the result by Bencze, Popovici and Smarandache [1]. More generally, D. Bode [2] defines m-perfect numbers as numbers n for which σ(σ . . . σ(n) . . . ) = 2n, {z } | m

and shows that for m ≥ 3 there are no even m-perfect numbers. The author [5] calls the superperfect numbers as σ ◦ σ-numbers. Let ψ(n) be the Dedekind function, ¶ Yµ 1 1+ ψ(n) = n , ψ(1) = 1 (p prime). p p/n

In our paper [5] many problems related to ψ ◦ ψ-perfect numbers and variants are considered. Among others there it is proved the following (see p.10) inequality: σ(σ(n)) ≥ ψ(σ(n)) ≥ 2n (1) for all even n, with equality only for n = 2k , where 2k+1 − 1 = prime. By this relation, we can reobtain Suryanarayana’s and Kanold’s result on the form of even superperfect numbers. A new proof of this is obtained in [6], too. But our aim in this note is to prove a part of OQ.113, namely that for all k ≥ 3 there are arbitrarily long sequences of consecutive integers n such that σ ◦ σ ◦ · · · ◦ σ(n) > 2n. | {z }

(2)

k

First remark, that when σ(n) is even, from (1) we get σ(σ(σ(n))) ≥ 2σ(n) ≥ 2(n + 1) > 2n and this can be continued, and thus (2) remains true for all k ≥ 3, when σ(n) is even. Now, we recall the well known fact (see [3]) that σ(n) is odd only when n = a2 or n = 2a2 (a ∈ N∗ ). We will show that there exist arbitrary long sequences with consecutive terms in which no term is of the form a2 or 2a2 . Let Pi (i = 1, m) be m distinct odd primes, where m is given. By the Chinese Remainder Theorem there exists x ∈ N with the property x ≡ Pi − i + 1

(mod Pi2 ), 115

i = 1, 2, . . . , m.

(3)

Then x, x + 1, . . . , x + m − 1 are m consecutive integers and no one can be of the form a2 or 2a2 since, on base of (3), xi − 1 = Pi2 ki + Pi ,

ki ∈ N

and this is divisible by Pi but not by Pi2 (Pi odd prime).

References [1] M. Bencze, F. Popovici, F. Smarandache, About very perfect numbers, Octogon Math. Mag., 5(1997), no.2, 53-54. [2] D. Bode, Uber eine Verallgemeinerung der Vollkommen Zahlen, Dissertation, Braunnschweig, 1971. [3] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, 4ed, Oxford Univ. Press, 1960. [4] H.J. Kanold, Uber ”Superperfect numbers”, Elem. Math. 24(1969), 6162. [5] J. Sándor, On the composition of some arithmetic functions, Studia Univ. Babe¸s-Bolyai, Math., XXXIV, 1, 1989, 7-14. [6] J. Sándor, On even perfect and superperfect numbers (Romanian), Sem. Did. Mat., 8(1992), 167-168. [7] D. Suryanarayana, Superperfect numbers, Elem. Math., 14(1969), 16-17. [8] D. Suryanarayana, There is no superperfect number of the form p2 α, Elem. Math., 28(1973), 148-150.

10

On Ruzsa’s lovely pairs

1. In paper [3] we have studied, among others, also the following problem of I.Z. Ruzsa [4]: Find all numbers a, b such that σ(a) = 2b,

σ(b) = 2a.

(1)

Ruzsa called these numbers (a, b) a lovely pair (or lovely numbers), and in an unpublished manuscript he showed that if a and b are both even, then a = 2k (2q+1 − 1), 116

b = 2q (2k+1 − 1)

(2)

where 2k+1 − 1 and 2q+1 − 1 are both primes. Sándor [3] defined a super-lovely pair by σ(σ(a)) = 2b, σ(σ(b)) = 2a (3) and proved that if a and b are both even, then a = b, i.e. a and b are superperfect numbers. (He also gave a new proof of (2)). Since in a recent O.Q. [6] a generalization of (1) to more numbers (and with p in place of 2) is considered, we shall present here our elementary method on the above stated results. 2. Let a = 2k A, b = 2a b, where k, q ≥ 1 and A, B are odd numbers. Since σ is a multiplicative function σ(a) = (2k+1 − 1)σ(A), etc., so one has (2k+1 − 1)σ(A) = 2q+1 B,

(2q+1 − 1)σ(B) = 2k+1 A.

Since 2k+1 − 1 divides 2q+1 B, and is odd, B is divisible by 2k+1 − 1, i.e. B = (2k+1 − 1)B 0 ; similarly A = (2q+1 − 1)A0 with A0 , B 0 odd numbers. One gets σ(A) = 2q+1 B 0 , σ(B) = 2k+1 A0 . (4) Now, using the inequality σ(mn) ≥ mσ(n) with equality only for m = 1 (see e.g. [1], [2]), one can write 2q+1 B ≥ (2k+1 −1)A0 ·2q+1 (where we have used also σ(m) ≥ m + 1 with equality only for m = prime), so we have B 0 ≥ A0 . The inequality B 0 ≥ A0 follows in the same manner. Thus A0 = B 0 , and (4) gives A = (2k+1 − 1)S, B = (2q+1 − 1)S with 2k+1 − 1 and 2q+1 − 1 both primes. Since σ(B) = 2k+1 S, σ(A) = 2q+1 S and σ((2q+1 − 1)S) ≥ S · 2q+1 with equalities, only if S = 1 and 2q+1 − 1 = prime, we get B = 2k+1 − 1, A = 2q+1 − 1. So (2) is proved. 3. Now let a = 2k A, b = 2q B be a solution of (3) with a, b both even. Then σ(a) = (2k+1 − 1)σ(A), σ(b) = (2q+1 − 1)σ(B) and (3) becomes σ((2k+1 − 1)σ(A)) = 2q+1 B,

σ((2q+1 − 1)σ(B)) = 2k+1 A.

(5)

By the above method, one can write successively 2q+1 B ≥ σ(A)2k+1 ≥ A · 2k+1 ,

2k+1 A ≥ σ(B) · 2q+1 ≥ B · 2q+1 ,

so 2q+1 B ≥ σ(A)2k+1 ≥ A · 2k+1 ≥ σ(B) · 2q+1 ≥ B · 2q+1 .

(6)

By (6) we must have equalities in all signs of inequalities, implying A = B = 1 and 2k+1 − 1 = prime, 2q+1 − 1 = prime. Therefore a = 2k , b = 2q with 2k+1 − 1 and 2q+1 − 1 both primes. But we must have k = q, so a = b, and these give the superperfect numbers of Suryanarayana and Kanold (see [2]). The unitary, etc. analogs of lovely numbers are introduced in [5]. 117

References [1] J. Sándor, On the composition of some arithmetic functions, Studia Univ. Babe¸s-Bolyai, 34(1989), no.1, 7-14. [2] J. Sándor, On an even perfect and superperfect number, Notes Number Theory Discr. Math., Sofia, 7(2001), no.1,4-5. [3] J. Sándor, Perfect Numbers: old and new issues, perspectives, Publ. Sapientia Foundation, Cluj, Romania, 2002, 1-90. [4] I.Z. Ruzsa, Personal communication to the author, may 2003, Math. Inst. Hungarian Academy of Sciences. [5] J. Sándor, The unitary, exponential, etc. analogs of Ruzsa’s lovely numbers, to appear. [6] M. Bencze, OQ.1273, Octogon Math. Mag., 11(2003), no.2, 850.

11

Completely d-perfect numbers

Let f : N∗ → N be a given arithmetic function. Recently, J.L. Pe [2] has called a number n to be f -perfect if we have: X i = n. (1) i|n,i 1) as r Y i=1

where n =

r Y

pαi i

=

r Y (αi + 1)(αi + 2) i=1

2

(= d∗ (n))

(3)

pαi i is the prime factorization of n. Clearly n = 1 is a solution

i=1

of n = d∗ (n), so we may assume n > 1. Now, if n is odd, then pi ≥ 3, and the following inequality is true. (α + 1)(α + 2) Lemma 1. pα ≥ for p ≥ 3, with equality only for p = 3 2 (α + 1)(α + 2) and α = 1. This follows at once from 3α ≥ , with equality for 3 α = 1 only (simple induction argument, which we omit here). Lemma 1 implies that the only odd solution of (3) is n = 3. If n is even, the things are more complicated. The following auxiliary results will be used: (α + 1)(α + 2) Lemma 2. For α ≥ 4, 2α > ; 2 3 for α ≥ 2, 3α > (α + 1)(α + 2). 3 3 Lemma 3. For all α ≥ 1, 5α > (α + 1)(α + 2). 4 119

Now, if n is even n =

2α1

r Y

pαi i with pi ≥ 3. When α1 = 1, equation (3)

i=1

becomes 3

2r Y (αi + 1)(αi + 2)

2

i=1

=2

r Y

pαi i .

(4)

i=1

Remark that the similar equation arrives, too when α1 = 2. Now, by the second inequality of Lemma 2, since p2 ≥ 3, we cannot have α2 = 2, since then clearly µ ¶r Y µ ¶r Y r r r Y 3 3 (αi + 1)(αi + 2) αi pi > (α + 1)(α + 2) = , 3 2 2 i=2

i=2

i=2

so (4) is impossible. When α2 = 2 on the other hand, with p2 = 3, r = 2 one obtains from (4) the equation 3 · 3 · 2 = 2 · 32 , which is a solution. Since α1 = 1, we have obtained n = 21 · 32 = 18, which is a d-perfect number. For α1 = 2 we obtain another solution n = 22 · 32 = 36. For α2 = 1, p2 = 3 equation (4) becomes: r r Y 3 Y (αi + 1)(αi + 2) , pi ≥ 5 (5) pαi i = 2 2 i=3

i=3

and by Lemma 3, this is impossible. Finally, when α1 = 3, one gets the equation 5

r Y (αi + 1)(αi + 2)

2

i=2

=4

r Y

pαi i .

(6)

i=2

5 3 > , by Lemma 3, this is impossible if pi ≥ 5. Now, when p2 = 3, 4 8 5 the inequality 3α > (α + 1)(α + 2) is true for α ≥ 2, so remains p2 = 3, 8 α2 = 1. But then (6) becomes Since

5

r Y (αi + 1)(αi + 2) i=3

2

=4

r Y

pαi i ,

pi ≥ 5

(7)

i=3

and by the above mentioned inequality cannot have solution. Therefore, the only even solution is n = 18, and this finishes the proof of the theorem.

References [1] Ch. Ashbacher, On numbers that are pseudo-Smarandache and Smarandache perfect, Smarandache Notions J., 14(2004), 40-42. 120

[2] J.L. Pe, On a generalization of perfect numbers, J. Recr. Math. (to appear). [3] J. Sándor, The star function of an arithmetic function, Octogon Math. Mag., 11(2003), no.2, 580-582. [4] J. Sándor, On completely f -perfect numbers (to appear). [5] J. Sándor, On (f, g)-perfect numbers, Octogon Math. Mag., 12(2004), no.2A, 539-542.

12

On completely f -perfect numbers

1. Let f : N∗ → N be a given arithmetic function. Recently, J. L. Pe [3] has called a number n to be f -perfect, if X f (i) = n, (1) i|n,i 2k. Then n cannot be a solution to equations (4) or (6). The number n is a solution of (5) iff n = 6. The only solution of X this type of equation (7) is n = 28. Proof. We first calculate S = S(n). Since the proper divisors of i|n,1 2k, on the other hand we have p(2k − k) > 2k(2k − 2) ≥ k(k + 1) by the inequality 2(2k − k) ≥ k + 1, i.e. p(2k

2k+1 ≥ 3k + 1,

k≥1

(9)

It is easy to verify by induction that (9) holds true for all k ≥ 1. Therefore, equation (40 ) is impossible. Remark. The solution n = 12 = 22 · 3 with p = 3, k = 2 doesn’t satisfy p > 2k. b) Similarly, for (50 ), by (8) we should have satisfied the inequality 2k p−1 ≤ kp + k(k + 1). Now, by p > 2k we get p(2k − k) > 2k(2k − k) > k(k + 1) + 1 ⇔ k(2k+1 − 3k − 1) ≥ 2. Now, the inequality 2k+1 ≥ 3k + 2, 124

k≥2

(10)

holds true. Thus for k ≥ 2 we cannot have a solution. For k = 1, however, by Theorem 2 we get the solution n = 21 · 3 when p = 3 > 2 · 1 = 2. c) For (60 ) remark, that similarly we must have 2k p − p ≤ k(k + 1) + kp, or p(2k − k − 1) ≤ k(k + 1). Now, by p > 2k, and the inequality 2k+1 > 3k + 3,

k≥3

(11)

it follows that p(2k − k − 1) > 2k(2k − k − 1) > k(k + 1). Thus we could have eventually k = 1 or k = 2. By Theorem 3 we cannot have solutions. d) The equation (70 ), by (8) implies 2k p − p − 1 ≤ k(k + 1) + kp so p(2k − k − 1) − 1 ≤ k(k + 1). Now, by p > 2k, and 2k(2k − k − 1) > k(k + 1) + 1 ⇔ k(2k+1 − 3k − 3) > 1, this is true by 2k+1 ≥ 3k + 4,

k ≥ 3,

(12)

so we could have eventually k = 1 or k = 2, i.e. n = 2p or n = 22 p. By Theorem 3 this is possible only when p = 7, when p > 2k, i.e. 7 > 4 is satisfied. Corollary. There are no ordinary even perfect numbers which are S-perfect or completely S-perfect in sense 1. The only even perfect number which is Sperfect in sense 2 is n = 6. The only even perfect number which is completely S-perfect in sense 2 is n = 28. Proof. Let n be an even perfect number. Then, by Euclid-Euler’s theorem, n can be written as n = 2k p, where p is a prime of the form p = 2k+1 − 1. Now, p > 2k is true, since 2k+1 > 2k + 1, k ≥ 1. This follows e.g. by induction, and we omit the details. Theorem 4 implies the corollary. 4. Finally note, that in paper [4] we have proved that the only completely d-perfect numbers are n = 1, 3, 18 and 36 (here d(n) is the number of distinct divisors of n).

References [1] Ch. Ashbacher, On numbers that are pseudo-Smarandache and Smarandache perfect, Smarandache Notions J., 14(2004), 40-42. [2] P. Gronas, The solution of the diophantine equation σ1 (n) = n, Smarandache Function J., 4-5(1994), 14-16. [3] J. L. Pe, On a generalization of perfect numbers, J. Recr. Math. (to appear). [4] J. Sándor, On completely d-perfect numbers, Octogon Math. Mag., 12(2004), no.1, 257-259; no.2A, 749-750. 125

13

On (f, g)-perfect numbers

1. Let f : N∗ → N be a given arithmetic function. Recently, J.L. Pe [4] has called a number n to be f -perfect if we have X f (i). (1) n= i|n,i6=n

For f = I (where I(n) = n for all n), we get the classical notion of a perfect number, since in this case (1) gives n = σ(n) − n, so σ(n) = 2n. In paper [8], we have called a number n completely f -perfect, if X n= f (i). (2) i|n

X Put f = I − ϕ, where ϕ is Euler’s totient. Then by Gauss identity i|nϕ(i) = n, (2) gives again σ(n) = 2n. Therefore, the completely I − ϕperfect numbers are the ordinary perfect numbers. In [8] we have studied certain special classes of S-perfect or completely S-perfect numbers, where S is the Smarandache function S(n) = min{k ∈ N : n|k!}. Assuming S(1) = 0, or S(1) = 1 we get distinct types of S-perfect numbers. In paper [10] we have proved that all completely d-perfect numbers are n = 1, 3, 18 and 36 (where d(n) is the number of all divisors of n). 2. A common generalization of f -perfect and completely f -perfect numbers is the following notion: Definition. Let f, g : N∗ → N be two given functions. A positive integer n is called (f, g)-perfect number, if X g(n) = f (i). (3) i|n

For g = I, (3) reduces to (2), while for g = I + f one obtains (1). Therefore the (f, I)-perfect numbers are the completely f -perfect, while the (f, I + f )perfect numbers, the f -perfect numbers. For f = I, g = 2I the ordinary perfect numbers are reobtained. For f = I, g = 2I − 1 the almost perfect numbers, while for f = i, g = 2I + 1, the quasi-perfect numbers. Letting f = ε, where ε(n) = 1 for all n, g = ϕ we get the (ε, ϕ)-perfect numbers, i.e. the solutions of the equation ϕ(n) = d(n). (4) 126

Theorem 1. The only (ε, ϕ)-perfect numbers are n = 1, 3, 8, 10, 18, 24, 30. Proof. All solutions of equation (4) are the above numbers, a result discovered by A.P. Minin in 1894 (see [1]). In fact, ϕ(n) > d(n) for all n > 30, see e.g. [5], [7]. For f = I, g = ϕd one obtains the equation σ(n) = ϕ(n)d(n).

(5)

In 1988 ([7]) we have proved that all odd solutions of (5) are n = 1, 3. All even solutions of the form n = 14k, where 7 - k are n = 14 and n = 42. In [9] we proved also that there are one even solutions n with 3 - n, ω(n) ≥ 3, and also 6|n and ω(n) ≥ 4. (Here ω(n) denotes the number of distinct divisors of n). Thus Theorem 2. The only odd (I, ϕd)-perfect numbers are n = 1 and 3. The numbers n = 12 and n = 42 are even (I, ϕd)-perfect numbers. There are no such even perfect numbers n with 3 - n, ω(n) ≥ 3 or 6|n, ω(n) ≥ 4. Let f = I, g = I + k, where k is fixed. Then, the equation σ(n) = n + k

(6)

can have at most a finite number of solution. For k = 2 the only solution is n = 4 (see [5]). Theorem 3. There are at most finitely many (I, I + k)-perfect numbers. The only (I, I + 2)-perfect numbers is n = 4. Let now f = ε (where ε(n) = 1 for all n), and g = σ − I. Then (3) gives the equation σ(n) − n = d(n). (7) The single solution of (7) is n = 4 (see [5], [7]), therefore: Theorem 4. The only (ε, σ − I)-perfect number is n = 4. Let f = I, g = I(ω + 1). Then (3) gives σ(n) = n(ω(n) + 1).

(8)

In [5], it is shown that σ(n) ≤ n(ω(n) + 1), with equality only for n = 1. Thus: Theorem 5. The only (I, I(ω + 1))-perfect number is n = 1. Let f = ε, g = 2ω . Then we get from (3): d(n) = 2ω(n) . 127

(9)

All solutions to this equation are n = 1 or n = squarefree (i.e. product of distinct primes), see [5]. We have proved: Theorem 6. The (ε, 2ω )-perfect numbers are n = 1 and all squarefree numbers. Remark. The same is true for the (ε, 2Ω )-perfect numbers, where Ω(n) denotes the total number of prime factors of n. Let f = I, g = kI − ϕ, where k > 1 is fixed. Then (3) gives the equation σ(n) = kn − ϕ(n)

(10)

studied by C.A. Nicol [3]. He proved that all solutions must satisfy ω(n) >

log(k − 1) , log 2

and conjectured that for k > 2 all solutions are even. For k > 2, n cannot be squarefree. Theorem 7. There are no (I, kI − ϕ)-perfect numbers n with ω(n) ≤

log(k − 1) . log 2

There are no squarefree (I, kI − ϕ)-perfect numbers for k > 2. Let now f = I, g = σ ◦ ϕ, where ◦ denotes composition. S.W. Golomb [2] gave certain particular solutions to σ(n) = σ(ϕ(n)),

(11)

giving Theorem 8. The number n = 1, 87, 362, 1257, 1798, 5002, 9374, are all (I, σ ◦ ϕ)-perfect numbers. Are there more, or eventually infinitely many? Let ψ denote the Dedekind arithmetic function given by ¶ Yµ 1 (p prime). ψ(1) = 1, ψ(n) = n = n 1+ p p|n

For f = 2ϕ, g = ψ ◦ σ one obtains from (3) the equation ψ(σ(n)) = 2n

(12)

which in [6] are called as ψ ◦ σ-perfect numbers. By using the result from [6], we state: 128

Theorem 9. All (2ϕ, ψ ◦ σ)-perfect numbers which are given by n = 2k , where 2k+1 − 1 is a prime. If n is an odd perfect number of this type, then n 6≡ −1 (mod 3), n 6≡ 7 (mod 12), n 6≡ −4 (mod 21), n 6≡ −10 (mod 21). n = 3 is an odd solution. When f = 2ϕ, g = ψ ◦ σ, one obtains the equation σ(ψ(n)) = 2n,

(13)

and one has (see [6]) Theorem 10. All (2ϕ, σ ◦ ψ)-perfect numbers n which are odd, satisfy the implication: σ(p + 1) > 2p ⇒ p - n (p prime). The only even perfect number of this type is n = 2. It is conjectured that there are no odd perfect numbers of this type.

References [1] L.E. Dickson, History of the theory of numbers, vol.1, Chelsea (original 1919). [2] S.W. Golomb, Equality among number-theoretic functions, Abstract 88211-16, Abstracts Amer. Math. Soc., 14(1993), 415-416. [3] C.A. Nicol, Some diophantine equations involving arithmetic functions, J. Math. Anal. Appl., 15(1966), 154-161. [4] J.L. Pe, On a generalization of perfect numbers, J. Recr. Math. (to appear). [5] J. Sándor, Some diophantine equations for particular arithmetic functions (Romanian), Sem. teoria struct., no.53, 1989, 1-10, Univ. of Timi¸soara, Romania. [6] J. Sándor, On the composition of some arithmetic functions, Studia Univ. Babe¸s-Bolyai, Math., 34(1989), no.1, 7-14. [7] J. Sándor, Geometric theorems, diophantine equations and arithmetic functions, American Research Press, Rehoboth, 2002. [8] J. Sándor, On completely f -perfect numbers, to appear. [9] J. Sándor, On the equation σ(n) = ϕ(n)d(n), to appear. [10] J. Sándor, On completely d-perfect numbers, Octogon Math. Mag., 12(2004), no.1, 257-259, no.2A, 749-750. 129

14

On abundant and deficient numbers

In a recent proposed problem, see [1], it is stated that the product of m (m ≥ 1) abundant numbers is also abundant, i.e. if a1 , . . . , am are all abundant, then a1 , . . . , am will be abundant, too. Recall that a number a is called abundant if σ(a) > 2a. We note here that the following much stronger result holds true: Theorem 1. Any multiple of an abundant number, is abundant, too. Proof. Let σ(a) > 2a. Then for any n > 1, σ(na) ≥ nσ(a) > n(2a) = 2na, by the following: Lemma. If x, y ≥ 1 are integers, then σ(xy) ≥ xσ(y). Proof. Let d be any divisor of y. Then xd will be a divisor of xy. Writing this for all divisors d, after addition, we get xσ(y) ≤ σ(xy). For the recent state of art on abundant numbers, see [2]. For example, for any k ≥ 8, the number n = 1 · 3 · 5 . . . (2k − 1) is abundant. For any k ≥ 1 fixed, there exist k consecutive abundant numbers. If a, b are given, then there exist infinitely many abundant integers n ≡ a (mod b). Every integer > 20162 can be expressed as the sum of two abundant numbers. There exist infinitely many sequences of five consecutive deficient numbers. For many other results (e.g. on primitive abundant, superabundant, highly abundant, etc. numbers), see [2] and the References therein. We now investigate some questions similar to Theorem 1 on deficient numbers (i.e. numbers a such that σ(a) < 2a). Clearly, any prime p is deficient, since σ(p) = p + 1 < 2p. We have: 4 Theorem 2. If the integer n satisfies σ(n) < n, then, if p - n, then np 3 will be deficient. Proof. Remark that σ(np) = σ(n)σ(p) for n > p, since p being a prime, p 2 3 4 (n, p) = 1. Now ≥ for p ≥ 2, so p + 1 ≤ p, i.e. σ(np) < n(p + 1) ≤ p+1 3 2 3 4 3 n p = 2np, so np will be deficient. 3 2 4 Remark. Let q > p be any prime q ≥ 5. Then σ(q) = q+1 < q ⇔ q > 3, 3 so Theorem 2 applies. Therefore there exist infinitely many multiples of a deficient number, which is deficient, too. On the other hand: Theorem 3. Let p be a prime, p - n, and suppose that n is abundant. Then np will be an abundant number. Proof. σ(np) = σ(n)σ(p)(p+1) > 2n(p+1) > 2np, so np will be abundant. 130

For example, if q is a prime and p > 2q, then n = 1 · 3 · 5 . . . (2q − 1) is such an abundant number. Therefore, there exist infinitely many multiples of a deficient number, which are not deficient. Theorem 4. If a, b satisfy the inequality σ(a)σ(b) < 2ab, then ab will be a deficient number. Proof. It is well-known that σ(ab) ≤ σ(a)σ(b), with equality only if (a, b) = 1. Since σ(a)σ(b) < 2ab, the result follows. For example, let p ≥ 5, q ≥ 2 be two primes. Then σ(p)σ(q) = (p + 1)(q + 1) < 2pq ⇔ pq + p + q + 1 < 2pq ⇔ (p − 1)(q − 1) > 2, i.e. (p − 1)(q − 1) ≥ 3. For p − 1 ≥ 3, q − 1 ≥ 1, this is true. For another example, let a = 2k , b = p (prime), where p > 2k+1 − 1. Then σ(a)σ(b) = (2k+1 − 1)(p + 1) < 2 · 2k p = 2k+1 p ⇔ 2k+1 p + 2k+1 − p − 1 < 2k+1 p ⇔ p > 2k+1 − 1. Since 2k is also deficient, we can state that there are infinitely many deficient numbers a, b such that ab is also deficient. On the other hand: Theorem 5. There exist infinitely many deficient numbers c, d such that cd is abundant. Proof. Let c = 2k , d = 3k , where k ≥ 2. Then c and d are deficient, since σ(c) = 2k+1 − 1 < 2 · 2k ,

σ(d) =

3k+1 − 1 < 2 · 3k , 2

by 4 · 3k > 3k+1 − 1 = 3 · 3k − 1. On the other hand, µ k+1

σ(cd) = σ(c)σ(d) = (2

− 1)

3k+1 − 1 2

¶ > 2 · 2k · 3k

since this is equivalent to 2k+1 · 3k+1 − 2k+1 − 3k+1 + 1 > 4 · 2k · 3k , i.e. 2 · 2k · 3k − 2 · 2k − 3 · 3k > −1. 131

By (2 · 2k − 3)(3k − 1) = 2 · 2k · 3k − 2 · 2k − 3 · 3k + 3, we have to prove that (2 · 2k − 3)(3k − 1) > 2. This is trivial, since 2 · 2k − 3 ≥ 2 · 22 − 3 = 5,

3k − 1 ≥ 32 − 1 = 8.

For the frequency of deficient numbers, see [3], [2].

References [1] M. Bencze, PP.5003, Octogon Math. Mag., 12(2004), no.1, 375. [2] J. Sándor, Abundant numbers, in M. Hazewinkel, Encyclopedia of Maths., supplement III, Kluwer Acad. Publ., 2001, 19-21. atai concerning the frequency [3] J. Sándor, On a method of Galambos and K´ of deficient numbers, Publ. Math., Debrecen, 39(1991), 155-157.

15

On abundant and nobly-abundant, or deficient numbers

1. Let d(n) and σ(n) denote the number, respectively - sum of divisors of a positive integer n. Recall that n is called abundant, if σ(n) > 2n, and deficient when σ(n) < 2n. The number n is called perfect, if σ(n) = 2n. In a recent paper, Jason Earls [1] posed some interesting problems on abundant and deficient numbers. A number n is called nobly-abundant when both of d(n) and σ(n) are abundant numbers. Similarly, n is called nobly-deficient, when d(n) and σ(n) are deficient (see Problems 13 and 14 of [1]). We wish to mention here that Kevin Ford [3] introduced the sublime numbers, which are numbers n such that both of d(n) and σ(n) are perfect. There are known up to now only two such numbers, namely n = 12, and n = 2126 (261 − 1)(231 − 1)(219 − 1)(27 − 1)(25 − 1)(23 − 1). Also, it is not known, if there exist odd sublime numbers. Our aim is to study certain properties of numbers n such that d(n) or σ(n) (or both) are abundant or deficient. 2. Let a be a given abundant number. Then all numbers n such that d(n) = a can be determined. (E.g. n = pa−1 , where p is an arbitrary prime.) Let e.g. a = 12. Then d(n) = 12 (1) 132

only if, with n = pa11 . . . par r (prime factorization) we have (a1 + 1) . . . (ar + 1) = 12, and since ai + 1 ≥ 2, this implies r ≤ 3. For r = 1 one has a1 = 11, so n = p11 1 ; when r = 2, (a1 + 1)(a2 + 1) = 12 gives a1 = 2, a2 = 3 or a1 = 1, a2 = 5, so n = p21 p32 or n = p1 p52 ; when r = 3, (a1 + 1)(a2 + 1)(a3 + 1) = 12 gives a1 = 1, a2 = 1, a3 = 2, so n = p1 p2 p23 . We will study only the last case, letting for simplicity n = pqr2 , where p, q, r are distinct primes. Since σ(n) = σ(pqr2 ) = (p + 1)(q + 1)(r2 + r + 1), the inequality σ(n) > 2n becomes r2 [(p + 1)(q + 1) − 2pq] + r(p + 1)(q + 1) + (p + 1)(q + 1) > 0

(2)

This is satisfies e.g. when p = 2, q = 3 (when (p + 1)(q + 1) − 2pq = 0), so we have proved: Theorem 1. Let r ≥ 5 be a prime. Then n = 2 · 3 · r2 has the property that both n and d(n) are abundant. For n = 2 · 3 · r2 = 6r2 we have σ(n) = 12(r2 + r + 1) > 12r2 . Let q = r2 + r + 1, and suppose that q is a prime. Since q > 3, clearly (q, 12) = 1, so σ(σ(n)) = σ(12)σ(q) = 28(q + 1) = 28(r2 + r + 2) > 24(r2 + r + 1) = 2σ(n). Thus: Theorem 2. Let r ≥ 5 be a prime, and suppose that q = r2 + r + 1 is a prime, too. Then n = 6r2 has the property that n, d(n), σ(n) are all abundant (i.e. n is an abundant and nobly-abundant number). Remark. Put r = 5, when q = 31 = prime. So n = 6 · 52 = 150 is an abundant and nobly-abundant number. For another example, let r = 17, when q = 307 = prime. So n = 6·172 = 1734 is also a nobly-abundant and abundant number. It seems very likely that there are infinitely many primes r such that q = r2 +r +1 is a prime, too. However, this seems very difficult to prove at present. Theorem 3. If r ≥ 5 is a prime, then n = 6r has the property that n is abundant, but d(n) is deficient. If p ≥ 2 is an arbitrary prime, then n = p is deficient, with d(n) deficient too. If p is a Mersenne prime, then σ(n) is deficient, too. Proof. d(6r) = 8, σ(8) = 15 < 16, σ(6r) = 12(r + 1) > 12r; d(p) = 2, σ(2) = 3 < 4, σ(p) = p + 1 < 2p. Also σ(σ(p)) = σ(p + 1) = σ(2k ), where p = 2k −1 is a Mersenne prime. Now σ(2k ) = 2k+1 −1, so σ(σ(p)) = 2k+1 −1 < 2k+1 = 2σ(p), i.e. n = p is deficient, and nobly-deficient, at the same time. Remark. Currently, there are known a number of 41 Mersenne primes, see e.g. [3]. 133

Theorem 4. Let p be a prime of the form p = 2k · N − 1, where N > 1 is odd and N < 2k+1 − 1. Then n = p has the property that n is deficient, d(n) is deficient, but σ(n) is abundant. Proof. σ(p) = p + 1 = 2k · N , so σ(σ(p)) = σ(2k · N ) = (2k+1 − 1)σ(N ) ≥ (2k+1 − 1)(N + 1) by σ(N ) ≥ N + 1 for N > 1. Now (2k+1 − 1)(N + 1) > 2σ(p) = 2k+1 · N ⇔ N < 2k+1 − 1, and the proof is complete. Remark. For example, p = 22 · 3 − 1 = 11 satisfies, with k = 2, N = 3 the required property. Another example is p = 22 · 5 − 1 = 19. We conjecture that there are infinitely many primes of this form.

References [1] J. Earls, Some Smarandache-type sequences and problems concerning abundant and deficient numbers, Smarandache Notions J., 14(2004), 243-250. [2] Josh Findley, The discovery of the 41th Mersenne prime, www/ams.org [3] K. Ford, Math Pages, http://www.mathpages.com/home/kmath2002/kmath2002.htm

16

The numbers n for which σ(n), d(n), ϕ(n) are abundant or deficient

1. Recently, Jason Earls [2] has considered certain problems on abundant and deficient numbers. Particularly, he proposed (see Problem 13) the study of ”nobly-abundant” numbers, i.e. numbers n with the property that σ(n) and d(n) are abundant at the same time. Recall that σ(n) and d(n) denote the sum, resp. number-of divisors of n. Let ϕ(n) denote the Euler totient, i.e. the cardinality of those x ≤ n with (x, n) = 1. A number n is called abundant, if σ(n) > 2n, and deficient if σ(n) < 2n. For nobly-abundant or deficient numbers, see another note by us [3]. 2. We now prove results, where appears also the abundancy or deficiency of ϕ(n). Theorem 1. There are infinitely many numbers n such that σ(n), d(n), and ϕ(n) are abundant at the same time. Proof. The proof is based on the following well-known lemma: 134

Lemma 1. Any multiple of an abundant number is abundant, too. Proof. Let m be abundant, i.e. σ(m) > 2m. Then k · m will be abundant, for any k ≥ 1, too. Indeed, if a|b, then σ(b) σ(a) X 1 X 1 = = ≤ , 0 a d d b 0 d|a

d |b

σ(m) σ(km) σ(km) ≤ , implying > 2, as desired. m km km Now, let p be a prime, 11 - p, and t be an integer such that (p, t) = 1, (t, 11) = 1 and 13|t. Put n = 11 · p11 · t (1) so by m|km we get

Then d(n) = 2 · 12 · d(t), σ(n) = 12 · σ(p11 ) · σ(t), ϕ(n) = 10 · ϕ(p11 ) · ϕ(t) = 10 · ϕ(p11 ) · 12 · K, by the multiplicative property of d, σ, ϕ and by 13|t ⇒ ϕ(13)|ϕ(t), i.e. ϕ(t) = 12 · K. Since 12 is an abundant number by the lemma, d(n), σ(n), ϕ(n) will be all abundant, too. This proves Theorem 1. Remarks. 1) For example, for p = 2, t = 13 we get the number n = 11 · 211 · 13 = 292864. For any multiple t of this number, for t odd, and (t, 11) = 1 we get other numbers. E.g. for t = 3 one gets n = 878592, etc. 2) For any t such that (t, 11) = (t, p) = (11, p) = 1 the numbers n of (1) have the property that d(n) and σ(n) are abundant (i.e. nobly-abundant numbers). For p = 2, t = 1 we get the number 11 · 211 = 22528. Theorem 2. There are infinitely many numbers n such that σ(n) is abundant, and ϕ(n) is deficient. Proof. First we show that: Lemma 2. When m ≡ 11 (mod 12), then σ(2m+1 − 1) ≥

5 · 2m+1 − 2 3

(2)

Proof. Since 22 ≡ 1 (mod 3), 24 ≡ 1 (mod 5), 23 ≡ 1 (mod 7), by 12|(m+ 1) we can write 2m+1 = 212 = (22 )6k ≡ 1 (mod 3). Similarly 2m+1 = (24 )3k ≡ 1 (mod 5), and 2m+1 = (23 )4k ≡ 1 (mod 7). Therefore, 3, 5, 7 are all prime factors of 2m+1 − 1. Thus 3, 5, 7, 3 · 5, 3 · 7, 5 · 7, 3 · 5 · 7 and 1, 2m+1 − 1 are all distinct divisors of 2m+1 − 1 ≥ 212 − 1 = 383; so µ ¶ X 1 1 2 σ(2m+1 − 1) = (2m+1 − 1) ≥ (2m+1 − 1) 1 + m+1 + , d 2 −1 3 m+1 d|(2

since

−1)

1 1 1 1 1 1 1 87 2 + + + + + + = > . 3 5 7 3·5 3·7 5·7 3·5·7 105 3 135

Now, m+1

(2

µ − 1) 1 +

2 + m+1 2 −1 3 1

¶ =

5 · 2m+1 − 2 , 3

and (2) follows. Now, put n = 3 · 2m , where m ≡ 11 (mod 12).

(3)

Since σ(n) = 4(2m+1 − 1), and σ(σ(n)) = 7σ(2m+1 − 1), by (2) one has σ(σ(n)) ≥ 7

5 · 2m+1 − 2 > 8(2m+1 − 1) = 2σ(n), 3

since this is equivalent to 11 · 2m+1 > −10. Thus σ(n), where n is given by (3), is abundant. On the other hand, it is easy to see that ϕ(n) is deficient, since ϕ(3 · 2m ) = ϕ(3) · ϕ(2m ) = 2 · 2m−1 = 2m , and σ(ϕ(3 · 2m )) = σ(2m ) = 2m+1 − 1 < 2ϕ(3 · 2m ) = 2m+1 . This finishes the proof of Theorem 2. Remark 3. Since d(3 · 2m ) = 2(m + 1) = 2 · 12 · h = 23 · 3 · h, by letting (h, 6) = 1 we get σ(d(3·2m )) = σ(23 )·σ(3)·σ(h) = 60σ(h) > 60h > 2(23 ·3·h) = 2d(3·2m ), we can say that the numbers 3·2m for m = 12h−1, where (h, 6) = 1, are nobly-abundant numbers. For example, for h = 1, m = 11 we get the number n = 3 · 211 = 6144. Theorem 3. For infinitely many n, ϕ(n) is abundant, while d(n) is deficient. 2 Proof. Let n = 3p −1 . Then µ ¶ 1 2 p2 −1 ϕ(n) = 3 1− = 2 · 3p −2 , 3 Ã σ(ϕ(n)) = 3

2 −1

3p

−1

2

! 2 −2

> 4 · 3p

,

a−3 4 2 2 > a, where a = 3p . So a > 27, i.e. 3p > 33 , and this is true for 2 9 any p ≥ 2. Thus ϕ(n) is abundant. Assume now that p is a prime. Then, since d(n) = p2 , we have σ(d(n)) = σ(p2 ) = p2 + p + 1 < 2d(n) = 2p2 , so d(n) will be deficient. if

136

Remark 4. By Remark 3 and Theorem 2, there are infinitely many n0 such that ϕ(n0 ) is deficient, and d(n0 ) is abundant. Finally, we prove Theorem 4. There are infinitely many n such that σ(n), d(n) and ϕ(n) are deficient at the same time. Proof. Let p be a prime, and n = 2p−1 . We will show that for sufficiently large p, this n will satisfy the required property. First remark that d(n) = p is clearly deficient, since σ(p) = p + 1 < 2p. On the other hand, ϕ(2p−1 ) = 2p−2 for p ≥ 2, so σ(ϕ(2p−1 )) = σ(2p−2 ) = 2p−1 − 1 < 2p−1 = 2ϕ(2p−1 ), implying that ϕ(n) is deficient, too. Finally, σ(2p−1 ) = 2p − 1. Now, recall a result of R. Bojanić [1]: σ(2p − 1) =1 p→∞ 2p − 1 lim

(4)

3 3 Thus, for p ≥ p0 , σ(2p − 1) < (2p − 1) = σ(2p−1 ), which is stronger than 2 2 σ(σ(2p−1 )) < 2σ(2p−1 ), implying the deficiency of σ(2p−1 ).

References [1] R. Bojanić, Asymptotic evaluations of the sum of divisors of certain numbers (Serbo-Croatian), Bull. Soc. Math.-Phys. R. P. Macédoine 5(1954), 5-15. [2] J. Earls, Some Smarandache-type sequences and problems concerning abundant and deficient numbers, Smarandache Notions J., 14(2004), 243-250. [3] J. Sándor, On numbers n for which σ(n) and d(n) are abundant, or deficient, submitted.

17

On S-abundant and deficient numbers

1. In a recent paper Jason Earls [2] has considered certain interesting problems and conjectures related to abundant or deficient numbers, involving Smarandache type notions. Recall that a number n is called abundant, if σ(n) > 2n, and deficient, if σ(n) < 2n, where σ(n) denotes the sum of divisors of n. Let S(n) be the Smarandache function, defined by S(n) = min{k ≥ 1 : n|k!} 137

(1)

The pseudo-Smarandache function is defined by ½ ¾ k(k + 1) Z(n) = min k ≥ 1 : n| 2

(2)

Problems 8 and 9 of [2] ask for a study of properties of numbers n for which S(n), resp. Z(n) are abundant. In what follows, these numbers will be called as S-abundant, resp. Z-abundant numbers, i.e. satisfying σ(S(n)) > 2S(n),

(3)

σ(Z(n)) > 2Z(n)

(4)

If the inequalities in (3), resp. (4) are reversed, then we use the S-deficient, resp. Z-deficient terminology. Clearly, in case of equalities, S(n), resp. Z(n) are perfect numbers. 2. Let p be a prime. Since S(p) = p, and σ(p) = p + 1 < 2p, clearly p is S-deficient. Similarly, if q is another prime, then since S(pq ) = p · q, for p < q we have σ(pq) = (p + 1)(q + 1) < 2pq since this is equivalent to p + q + 1 < pq, i.e. (p − 1)(q − 1) > 2. If p = 2, q = 3, this is not true, but for p = 2, q ≥ 5 it is valid; and also for 3 ≤ p < q. For p = q, σ(pq) = σ(p2 ) = p2 + p + 1 < 2p2 by p + 1 < p2 . Therefore, we have proved: Theorem 1. There are no S-abundant numbers of the form n = pq , where p, q are primes. The only S-perfect numbers of this form are n = 23 and n = 32 . All other are S-deficient. The proof of Theorem 1 shows that the result can be extended to any n with S(n) = p · q, where p and q are primes. (For example, S(18) = 6 = 2 · 3, and 18 6= 23 or 32 ). Clearly, p and q are deficient numbers. On the other hand, one has: Theorem 2. If S(n) = a · b, where at least one of a and b is abundant, then n is S-abundant number. Proof. The proof is based on the well-known inequality σ(ab) ≥ aσ(b)

(5)

for any a, b ≥ 1. Now, suppose that b is abundant. Since σ(S(n)) = σ(ab) ≥ aσ(b) > a · 2b = 2ab = 2S(n), the theorem is proved. Remark. In [1] it is shown that for an arbitrary prime p, and any integer n ≥ 1, it is possible to find a number k such that S(pk ) = n · p 138

(6)

Now, let n be abundant. By Theorem 2, (6) shows that for any prime p there exists a positive integer k such that pk is S-abundant. It is well known that Z(p) = p − 1 (p = prime). Let p − 1 = 2k · N . If N = 1, then p = 2k + 1 = Fermat prime. Then σ(Z(p)) = σ(2k ) = 2k+1 − 1 < 2 · 2k = 2Z(p). Thus, any Fermat prime is Z-deficient. On the other hand: Theorem 3. Let p be a prime of the form p = 2k · N + 1, where N > 1 is odd, and suppose that N < 2k+1 − 1. Then p is Z-abundant number. Proof. σ(Z(p)) = σ(p − 1) = σ(2k N ) = σ(2k )σ(N ) ≥ (2k+1 − 1)(N + 1) > 2(2k · N ) = 2Z(p) ⇔ 2k+1 · N + 2k+1 − N − 1 > 2k+1 · N ⇔ N < 2k+1 − 1. For example, 13 = 22 · 3 + 1, where k = 2, N = 3, so 3 < 23 − 1; thus n = 13 is Z-abundant. Another example is n = 41 = 23 · 5 + 1. Finally, we prove Theorem 4. Let k be even. Then n = 2k · 3 is a Z-deficient number. 2k+1 (2k+1 + 1) , and 2k+1 is the smallest Proof. If k is even, then 2k · 3| 2 power of 2 with this property, so Z(2k · 3) = 2k+1 . But 2k+1 is deficient, since σ(2k+1 ) = 2k+2 − 1 < 2 · 2k+1 , and the result follows.

References [1] Ch. Ashbacher, An introduction to the Smarandache function, Erhus. Univ. Press, Vail, 1995. [2] J. Earls, Some Smarandache-type sequences and problems concerning abundant and deficient numbers, Smarandache Notions J., 14(2004), 243-250.

18

The square deficiency of a number

In [1] the deficiency of a number n is denoted by α(n) = 2n − σ(n), where σ(n) denotes the sum of divisors of n. Problem 18 of [1] asks for numbers n with α(n) a perfect square. It is wrongly stated that n = 2597 is the least nonprime such that α(n) is a square. Indeed, it is easy to see that for n = 14 one has σ(14) = 24, so α(14) = 4 = 22 . In fact one has, more generally: Theorem 1. Let p be a prime of the form p = A2 + 3. Then α(2 · p) = square. 139

Proof. Let n = q · p, where p, q are distinct primes. Then α(n) = 2qp − σ(qp) = pq − p − q − 1. For q = 2, α(2p) = p − 3 = A2 for p = A2 + 3. Remark 1. Probably, there exist infinitely many primes in the sequence {A2 + 3}. However, this seems to be a very difficult problem. For example, one obtains primes for A = 2, 4, 8, 10, 14, 28, 38, 50, . . . For A = 50 one has the prime p = 2503, so n = 2p = 5006, which is greater than 2597. Theorem 2. Let p be a prime of the form p = A2 +1. Then α(p) = square. Proof. α(p) = 2p − σ(p) = 2p − (p + 1) = p − 1 = A2 . Remark 2. As above, one can conjecture that there exist infinitely many primes of the form A2 + 1. One obtains primes for A = 1 , 2, 4, 6, 10, 14, 16, 20, 24, 26, 36,. . . giving the numbers n = 2, 5, 17, 37, 101, 197, 257, 401, 577, 677, 1297,. . . Theorem 3. Let p, q be distinct primes such that p2 q − p2 − pq − p − q = A2 + 1

(1)

Then α(p2 q) = square. Proof. α(p2 q) = 2p2 q − σ(p2 q) = 2p2 q − (p2 + p + 1)(q + 1) = p2 q − p2 − pq − p − q − 1 = A2 . Remark 3. Let q = 2. Then condition (1) becomes p2 − 3p − 3 = A2 . This is satisfied by p = 7, since 49 − 24 = 52 . Thus n = 72 · 2 = 98 has a square deficiency, too. For p = 7, relation (1) becomes 41q − 57 = A2

(2)

The least prime q for which this is solvable if q = 53, when A = 46. Then n = 72 · 53 = 2597. Theorem 4. Let p be a prime of the form p = A2 + 2k+1 − 1 (A ≥ 1, k ≥ 1 integers). Then α(2k · p) = square. Proof. α(2k · p) = 2k+1 · p − (2k+1 − 1)(p + 1) = p + 1 − 2k+1 = A2 . Remark 4. Put A = 2, when p = 2k+1 + 3. This is prime for k = 1, 2, 3, 5, 6, 11, . . . when p = 7, 11, 19, 67, 131, 4099, . . . giving the numbers with the same square deficiency 4 : n = 14, 44, 152, 2144, 8384, 8394752, . . . We can conjecture that there are infinitely many primes p of this form. For A = 4 we get p = 2k+1 + 15, which is prime for k = 1, 2, 3, . . . giving p = 19, 23, 31, . . . , when n = 38, 92, 248, . . . All these numbers have the same deficiency α(n) = 16. 140

References [1] J. Earls, Some Smarandache-type sequences and problems concerning abundant and deficient numbers, Smarandache Notions J., 14(2004), 243-250.

19

Exponentially harmonic numbers

1. Let σ(n) and d(n) denote the sum, resp. number of divisors of n. In 1948 O. Ore [12] called a number n harmonic if σ(n)|nd(n)

(1)

See e.g. G. L. Cohen and R. M. Sorli [2] for such numbers. If σk (n) denotes the sum of kth powers of divisors of n (k ≥ 1 integer), then G. L. Cohen and M. Deng [3] introduced k-harmonic numbers by σk (n)|nd(n)

(2)

A perfect number is always harmonic (see [12]), and Ore conjectured that all harmonic numbers are even. This is a quite deep conjecture, since, if true, clearly would imply the non-existence of odd perfect ³ numbers. n´ A divisor d of n is called unitary divisor if d, = 1. If σ ∗ (n), d∗ (n) d are the sum, resp. number of unitary divisors of n, then n is called unitary harmonic if σ ∗ (n)|nd∗ (n), (3) see K. Nageswara Rao [11], P. Hagis and G. Lord [5], Ch. Wall [19]. A divisor d of n is called a bi-unitary divisor, if the greatest common n unitary divisor of d and is 1. If σ ∗∗ (n), d∗∗ (n) are the sum, and number d of bi-unitary divisors of n, recently we have introduced (see [16]) bi-unitary harmonic numbers by σ ∗∗ (n)|nd∗∗ (n) (4) For infinitary harmonic numbers, related to the concept of an ”infinitary divisor”, see P. Hagis and G. L. Cohen [7]. Let n > 1 be a positive integer having the prime factorization n = pa11 . . . par r . A divisor d of n is called exponential divisor, if d = pb11 . . . pbrr where b1 |a1 , . . . , br |ar . This notion is due to E. G. Straus and M. V. Subbarao [17]. Let σe (n) and de (n) be the sum and number of, exponential divisors of 141

n. Let by convention σe (1) = de (1) = 1. Straus and Subbarao have introduced e-perfect numbers n by σe (n) = 2n

(5)

They proved the non-existence of odd e-perfect numbers, with related other results. For results on e-superperfect numbers (i.e. satisfying σe (σe (n)) = 2n), see [8]. For density problems, e-perfect numbers not divisible by 3, or emultiperfect numbers, see P. Hagis [6], L. Lucht [10], J. Fabrykowski and M. V. Subbarao [4], W. Aiello et al. [1]. For results on de (n), we quote J. M. DeKoninck and A. Ivić [9]. For the exponential totient function ϕe (n), see J. Sándor [13]. For e-convolution and a survey connected to the Möbius function, see J. Sándor and A. Bege [14]. For multiplicatively e-perfect numbers, see J. Sándor [15]. 2. The aim of this note is study two notions of e-harmonic numbers. An integer n will be called e-harmonic of type 1 if σe (n)|nde (n)

(6)

In all examples of section 1 the harmonic numbers notions were suggested by the consideration of the harmonic means of the considered divisors. For example, if 1 = d1 < d2 < · · · < dr = n are all divisors of n, then their harmonic mean is ¶ µ 1 1 + ··· + . H(n) = r/ d1 dr Since

X 1 1X n 1X σ(n) = = dr = , dr n dr n n

we get H(n) =

nd(n) , σ(n)

(7)

so a harmonic number n is a number such that H(n) is an integer. E.g. for H(n) = 2 we get the so-called ”balanced numbers” proposed by M. V. Subbarao [18], with single solution n = 6. Now, if we consider the harmonic mean He (n) of the exponential divisors (e) (e) d1 , . . . , dr , then r , He (n) = X 1 (e)

di 142

where r = de (n) denotes the number of exponential (or e-) divisors of n. Let pa be a prime power. Then the e-divisors of pa are pd with d|a, so X 1

=

(e)

di

X 1 1 X a−d = p pa pa d|a

d|a

in this case. When n = pa q b (p 6= q primes) one obtains similarly    X 1 X X X 1 1  1  = = d1 q d2 d1 d2 (e) p p q d i

d1 |a,d2 |b

d1 |a

 =

1 

pa q b

X

d2 |b

  X pa−d1   q b−d2  .

d1 |a

d2 |b

In the general case, when n = pa11 pa22 . . . par r , one has He (n) = where Se (n) =

r Y i=1

nde (n) , Se (n) 

 

(8)

X

pai i −di 

(9)

di |ai

We say that n is e-harmonic of type 2 if He (n) is integer, i.e. Se (n)|nde (n),

(10)

where Se (n) is given by (9). Theorem 1. If n is squarefree, then it is e-harmonic of both types. Proof. If n is squarefree, i.e. n = p1 p2 . . . pr , then clearly by definitions  of σe (n)and Se (n) one has σe (n) = p1 p2 . . . pr = n and Se (n) = r Y X 1−d  pi i  = 1, so (6) and (8) are satisfied. We shall see later (see the i=1

di |1

Remark after Theorem 3), that there exist also numbers with this property, which are not squarefree. Theorem 2. Let n = pa11 . . . par r be the prime factorization of n > 1. If n is e-perfect, then n is e-harmonic of type 1 if and only if at least one of a1 , . . . , ar is not a perfect square. 143

Proof. If σe (n) = 2n, then (6) gives 2|de (n). It is well-known that de (n) = d(a1 ) . . . d(ar ), so at least one of d(a1 ), . . . , d(ar ) must be even. But, it is wellknown that d(a) is even iff a is not a perfect square, so the result follows. Remark. 1) Since e.g. 22 ·33 ·52 , 23 ·32 ·52 , 24 ·33 ·52 ·112 , 26 ·33 ·52 ·72 ·132 , 27 · 32 · 52 · 72 · 132 are e-perfect numbers, by the above theorem, these are also e-harmonic numbers of type 1. 2) Similarly, if n is e − k perfect, i.e. σe (n) = kn (k ≥ 2 integer), see [1], then n is e-harmonic of type 1 iff k|d(a1 )d(a2 ) . . . d(ar )

(11)

In what follows we shall introduce another new notion. We say that n is modified e-perfect number, if Se (n)|n,

(12)

where Se (n) is given by (9). Since Se (p) = 1, Se (p2 ) = p+1, Se (p3 ) = p2 +1, we have Se (22 · 32 ) = (2 + 1)(3 + 1) = 22 · 3, Se (22 · 33 · 52 ) = (2 + 1)(32 + 1)(5 + 1) = 22 · 32 · 5, Se (23 · 32 · 52 ) = (22 + 1)(3 + 1)(5 + 1) = 23 · 3 · 5, so 22 · 32 , 22 · 33 · 52 , 23 · 32 · 52 (which are also e-perfect) are modified e-perfect numbers. Clearly n = 24 · 32 · 112 (which is e-perfect) is not modified e-perfect, since by Se (p4 ) = 1 + p2 + p3 , we have Se (24 ) = 1 + 22 + 23 = 13. Theorem 3. If n is modified e-perfect, then it is harmonic of type 2. Proof. This follows at once from (12) and (8). Remark. Thus 22 · 33 · 52 and 23 · 32 · 52 are e-harmonic numbers of both types (though they are not squarefree).

References [1] W. Aiello et al., On the existence of e-multiperfect numbers, Fib. Quart. 25(1987), 65-71. [2] G. L. Cohen and R. M. Sorli, Harmonic seeds, Fib. Quart. 36(1998), 386-390. [3] G. L. Cohen and M. Deng, On a generalization of Ore’s harmonic numbers, Nieuw Arch. Wiskunde 16(1998), no. 3, 161-172. [4] J. Fabrykowski and M. V. Subbarao, On e-perfect numbers not divisible by 3, Nieuw Arch. Wiskunde (4)4(1986), 165-173. [5] P. Hagis and G. Lord, Unitary harmonic numbers, Proc. Amer. Math. Soc. 51(1975), 1-7. 144

[6] P. Hagis, Some results concerning exponential divisors, Int. J. Math. Math. Sci. 11(1988), 3434-349. [7] P. Hagis and G. L. Cohen, Infinitary harmonic numbers, Bull. Austral. Math. Soc. 41(1990), 151-158. [8] J. Hanumanthachari et al., On e-perfect numbers, Math. Student, 46(1978), 71-80. [9] J. M. DeKoninck and A. Ivić, An asymptotic formula for reciprocals of logarithms of certain multiplicative functions, Canad. Math. Bull. 21(1978), 409-413. [10] L. Lucht, On the sum of exponential divisors and its iterates, Arch. Math. (Basel), 27(1976), 383-386. [11] K. Nageswara Rao, On some unitary divisor functions, Scripta Math., 28(1967), 347-351. [12] O. Ore, On the averages of the divisors of a number, Amer. Math. Monthly, 55(1948), 615-619. [13] J. Sándor, On an exponential totient function, Studia Univ. Babe¸s-Bolyai Math. 41(1996), 91-94. obius function: generalizations and exten[14] J. Sándor and A. Bege, The M¨ sions, Adv. Stud. Contemp. Math. 6(2003), no. 2, 77-128. [15] J. Sándor, On multiplicatively e-perfect numbers, (to appear). [16] J. Sándor, Bi-unitary harmonic numbers, (to appear). [17] E. G. Straus and M. V. Subbarao, On exponential divisors, Duke Math. J. 41(1974), 465-471. [18] M. V. Subbarao, Problem E1558, Amer. Math. Monthly 70(1963), 92, Solution in 70(1963), 1009-1010. [19] Ch. R. Wall, Unitary harmonic numbers, Fib. Quart. 21(1983), 18-25.

145

146

Chapter 4

Algebraic and analytic inequalities ”... Our subject is difficult to define precisely, but belong partly to ’algebra’ and partly to ’analysis’.” (G.H. Hardy, J.E. Littlewood and G. Pólya)

”... Today inequalities play a significant role in all fields of mathematics, and they present a very active and attractive field of research.” (D.S. Mitrinović)

147

1

A monotonicity property of a product of sums of powers

Let pk > 0, xk > 0, k = 1, 2, . . . , n. We will prove that f (α) ≥ f (β) if α ≥ β > 0, where Ã n !Ã n ! X X α α f (α) = pk xk pk /xk . k=1

k=1

For simplicity, we take n = 3, x1 = x, x2 = y, x3 = z, p1 = p, p2 = q, p3 = r (the general case will follow on the same lines). Then µ ¶ p q r α α α f (α) = (px + qy + rz ) + + = p2 + q 2 + r 2 xα y α z α ·³ ´ µ ¶α ¸ ·µ ¶α ³ ´ ¸ h³ x ´α ³ z ´α i y α y α z x + rp + + qr + + . +pq y x z y z x Let

where a =

µ ¶α ³ ´ x y α g(α) = + = aα + a−α , y x x . Then y µ ¶ 1 aα + 1 α g (α) = ln a a − α = (ln a)(aα − 1). a aα 0

Now, remark that for all a > 0, α > 0 one has (ln a)(aα − 1) ≥ 0 (indeed, when α ≥ 1, then clearly ln a ≥ 0, aα ≥ 1, and when 0 < a < 1, then ln a < 0, aα < 1). Therefore, f 0 (α) ≥ 0 (being the sum of terms of the forms Ag 0 (α), with A ≥ 0), implying that f is an increasing function of α. The general case is exactly the same.

2

An application of the Cauchy-Bunjakovski inequality Theorem. Let xi , yi , zi be real numbers. If xi yi > zi2 (i = 1, n), then: v ! Ã n ! Ã n !2 uÃ n n q u X X X X t xi yi − zi . xi yi − zi2 ≤ (1) i=1

i=1

148

i=1

i=1

We will use the following Lemma. If ai , bi ≥ 0, then n p X

vÃ u n !Ã n ! X u X bi . ai ai bi ≤ t

(2)

i=1

i=1

i=1

Proof. By the Cauchy-Bunjakovski inequality one has Ã

n X √ p ai bi

!2 ≤

Ã n X

i=1

!Ã n ! X ai bi ,

i=1

and (2) follows. Now, let ui =

i=1

√ xi yi , and put ai = ui − zi , bi = ui + zi . Then

n q n q X X xi yi − zi2 = u2i − zi2 i=1

i=1

vÃ v !2 Ã n !2 u n u n n u X X X uX ≤ t (ui − zi ) (ui + zi ) = t ui − zi . i=1

i=1

i=1

i=1

1 √ Now, ui = ui √ yi , so again by the Cauchy-Bunjakovski inequality yi Ã n X

!2 ui

i=1

Ã n !Ã n ! Ã n !Ã n ! X u2 X X X i ≤ yi = xi yi , yi i=1

i=1

i=1

i=1

and (1) follows. We have proved vÃ !2 Ã n !2 u n n q u X√ X X xi yi − zi2 ≤ t xi yi − zi i=1

i=1

i=1

vÃ ! Ã n ! Ã n !2 u n u X X X ≤t xi yi − zi , i=1

i=1

which is stronger then (1). 149

i=1

(3)

3

The Chrystal inequality and its applications to convexity Let f (x) = ln(1 + ex ), x ∈ R. Then, since f 00 (x) =

ex > 0, (1 + ex )2

f is strictly convex on R. By Jensen’s inequality one can write f (α1 x1 + · · · + αn xn ) ≤ α1 f (x1 ) + · · · + αn f (xn ) for all xi ∈ R, αi > 0,

n X

αi = 1, i = 1, n. Thus

i=1

ln(1 + eα1 x1 eα2 x2 . . . eαn xn ) ≤ ln(1 + ex1 )α1 . . . (1 + exn )αn . Let exi = yi > 0. One obtains 1 + y1α1 . . . ynαn ≤ (1 + y1 )α1 . . . (1 + yn )αn ,

yi > 0.

(1)

bi , i = 1, n in (1). Then one obtains ai n X Theorem 1. For all ai , bi > 0, αi > 0, αi = 1, i = 1, n, one has Put yi =

i=1

aα1 1 . . . aαnn + bα1 1 . . . bαnn ≤ (b1 + a1 )α1 . . . (bn + an )αn .

(2)

1 one can write n p p √ n a1 . . . an + n b1 . . . bn ≤ n (b1 + a1 ) . . . (bn + an )

(3)

Remark. For αi =

which is known as the ”Chrystal inequality” (see [1]), when ai ≡ 1. Therefore, Theorem 1 can be regarded as the ”extended Chrystal inequality”. We now show how can be applied (2) to deduce in a unitary way certain convexity results. The following result appears also in [3]. Theorem 2. The geometric mean of positive concave functions is concave, too. Proof. Let fi , i = 1, n be positive concave functions. Put F (x) = (f1 (x))α1 . . . (fn (x))αn , 150

where αi > 0,

n X

αi = 1. Then

i=1

F (αx + βy) =

n n Y Y [fi (αx + βy)]αi ≥ [αfi (x) + βfi (y)]αi | {z } | {z } i=1

≥

i=1

n Y [αfi (x)]αi + [βfi (y)]αi = αF (x) + βF (y),

ai

bi

α, β > 0, α + β = 1.

i=1

Here we have applied (2) for ai = αfi (x), bi = βfi (y). Therefore fi concave (i = 1, n) ⇒ F concave functions. We now consider log-convex functions. Theorem 3. The sum of two positive log-convex functions is log-convex, too. Proof. Let f1 , f2 > 0 be log-convex, i.e. fi (αx + βy) ≤ (fi (x))α (fi (y))β , x, y ∈ I, α, β > 0, α + β = 1, fi : I → R, i = 1, 2. Then f1 (αx + βy) + f2 (αx + βy) ≤ (f1 (x))α (f1 (y))β + (f2 (x))α (f2 (y))β ≤ (f1 (x) + f2 (x))α (f1 (y) + f2 (y))β . Indeed, apply the Theorem for n = 2, α1 = α, α2 = β, a1 = f1 (x), a2 = f1 (y), b1 = f2 (x), b2 = f2 (y). The result follows. More complicated proof appears e.g. in [2]. Remark. It is well-known that a log-convex function is convex, too (but not in reciprocal order!) see e.g. [1], [2]. Corollary. If fi , i = 1, n, are log-convex functions, then f1 + · · · + fn is log-convex, too. This follows from Theorem 3, by induction. 1 Γ(x) Example. Since ln Γ(x) and ln are convex, we get that is logx x convex for x > 0.

References [1] G.H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Oxford, 1934. [2] A.W. Roberts, D.E. Varberg, Convex functions, Academic Press, 1973. [3] J. Sándor, Two theorems on convexity or concavity of functions, Octogon Math. Mag., 8(2000), no.2, 337-339. 151

Note added in proof. Some authors (including the present one) have attributed inequality (3) (when ai ≡ 1) to Huygens. However, it is clear from the monograph [1] (see p.61, Theorem 64) that this particular inequality appeared in the book ”Algebra” by G. Chrystal from 1900.

4

On certain inequalities for square roots and n-th roots

ˇ Arslanagić [1], remarks that the following inequality 1. In a recent note S. for two square roots is valid: p √ √ xy ≥ x − 1 + y − 1, x, y ≥ 1. (1) Then he applies relation (1) to three square roots, by showing that p p √ √ z(xy + 1) ≥ x − 1 + y − 1 + z − 1, x, y, z ≥ 1.

(2)

While the nice inequality (1) contains similar terms on both sides, its extension (2) is of another nature. Our desire would be to have an inequality of the following type: p √ √ √ xyz ≥ x − 1 + y − 1 + z − 1. (3) But this cannot be true for all x, y, z ≥ 1. Put e.g. z = 1, when (3) becomes p √ √ xy ≥ 1 + x − 1 + y − 1, x, y ≥ 1, (4) which is not generally true. But perhaps, for greater values of x, y, z, relation (3) holds true. The aim of this note is to consider such inequalities, along with extensions for n-th roots (n ≥ 2). 2. A refinement of (1) for greater values of x, y can be deduced from the following: Lemma 1. If a, b ≥ 1, then p √ √ √ (a + 1)(b + 1) ≥ 1 + ab ≥ a + b. (5) Proof. The first inequality of (5) is valid for all a, b > 0; in fact it is a particular case of Chrystal inequality (see [2]) p √ n (a1 + 1) . . . (an + 1) ≥ 1 + n a1 . . . an . (6) 152

The second relation of (5) follows from √ √ √ √ √ (1 + ab)2 = 1 + 2 ab + ab ≥ a + 2 ab + b = ( a + b)2 ⇔ ab − a − b + 1 ≥ 0 ⇔ (a − 1)(b − 1) ≥ 0. Put now a = x − 1, b = y − 1 for x, y ≥ 2. Then one gets: Theorem 1. For all x, y ≥ 2 one has p p √ √ xy ≥ 1 + (x − 1)(y − 1) ≥ x − 1 + y − 1.

(7)

Remark. Since (5) holds also for all a, b > 0 and (a − 1)(b − 1) ≥ 0, clearly (7) is valid also for x, y ∈ [1, 2]. Apply now the Cauchy-Bunjakovski inequality in order to deduce: Ã n Xp

!2 ai bi

Ã ≤

i=1

n X i=1

so

!Ã ai

n X

! bi

,

i=1

v v u n u n uX uX t ai bi ≤ ai · t bi .

n p X i=1

i=1

(8)

i=1

Now, let first n = 2, a1 b1 = a, a2 b2 = b. Then (8) implies s µ ¶ √ √ a b a + b ≤ (a1 + a2 ) + . a1 a2 Here

µ (a1 + a2 )

a b + a1 a2

¶ =a+b+b

a1 a2 +a . a2 a1

a2 b = µ, so one considers aµ + ≤ ab + 1 (where a + b + ab + 1 = a1 µ 2 (a + 1)(b + 1)). Now, this is equivalent· to the ¸ quadratic inequality aµ − (ab + 1 1)µ + b ≤ 0, having as solutions µ ∈ , b , (if ab ≥ 1). Therefore, if a, b > 0, a · ¸ 1 ab ≥ 1, µ ∈ , b , then a s p √ √ b a + b ≤ aµ + + a + b ≤ (a + 1)(b + 1) (9) µ Put

153

and this gives another refinement of type (5), with possible application of type (7). Let now b1 = · · · = bn = 1 in (8). Therefore p √ √ a1 + · · · + an ≤ n(a1 + · · · + an ). Now, if then clearly

n(a1 + · · · + an ) ≤ (a1 + 1) . . . (an + 1)

(10)

p √ √ a1 + · · · + an ≤ (a1 + 1) . . . (an + 1),

(11)

giving Theorem 2. If xi ≥ 1, i = 1, n, satisfy the inequality

then

n(x1 + · · · + xn − n) ≤ x1 . . . xn ,

(12)

√ √ √ x1 . . . xn ≥ x1 − 1 + · · · + xn − 1.

(13)

Proof. Apply (11) with (10) to ai = xi − 1. Remark. To obtain a characterization of an inequality p √ √ √ a + b + c ≤ (a + 1)(b + 1)(c + 1),

(14)

remark that after elementary computations, this becomes equivalent to X √ abc + (ab − 2 ab) + 1 ≥ 0. (15) √ √ Since ab−2 ab+1 = ( ab−1)2 ≥ 0, clearly this is satisfied e.g. if abc ≥ 2. Thus e.g. if abc ≥ 2. Thus e.g. if a ≥ 2, b ≥ 1, c ≥ 1, relation (14) holds true. Therefore e.g. if x ≥ 3, y ≥ 2, z ≥ 2, then p √ √ √ x − 1 + y − 1 + z − 1 < xyz. (16) Now, apply inequality (6). Searching for a relation of type √ √ √ 1 + n a1 . . . an ≥ n a1 + · · · + n an , √ put n ai = xi . Thus, an inequality 1 + x1 . . . xn ≥ x1 + · · · + xn should be proved. We have: Lemma 2. If x1 , x2 ≥ 2, xi ≥ 1, i ≥ 3, then 1 + x1 . . . xn ≥ x1 + · · · + xn , 154

n≥3

(17)

with equality only for n = 3, and x1 , x2 = 2, x3 = 1. Proof. 1 + x1 x2 x3 ≥ x1 + x2 + x3 can be written also as x1 (x2 x3 − 1) ≥ x2 + x3 − 1. Now, 1+x2 x3 ≥ x2 +x3 , since (x2 −1)(x3 −1) ≥ 0, so by putting x2 +x3 = a, by x1 ≥ 2 we get x1 (x2 x3 − 1) ≥ 2(a − 2) ≥ a − 1 ⇔ a ≥ 3. Now x2 + x3 = a ≥ 3 is true by assumption. The general case follows at once by mathematical induction. Theorem 3. For n ≥ 3, x1 ≥ 2n + 1, x2 ≥ 2n + 1, xi ≥ 1, i ≥ 3, one has p √ √ √ n x1 x2 . . . xn ≥ 1 + n (x1 − 1) . . . (xn − 1) > n x1 − 1 + · · · + n xn − 1. (18) Proof. From Lemma 2 it follows that for a1 , a2 ≥ 2n , ai ≥ 1, i ≥ 3 one has

p n

(1 + a1 ) . . . (1 + an ) >

√ √ n a1 + · · · + n an ,

(19)

which is obtained via inequality (6). By letting ai = xi −1, relation (18) follows. We cannot have equality (even for n = 3), since in (6) there is equality only for a1 = · · · = an , and this is not possible in (19) by Lemma 2.

References ˇ Arslanagić, A refinement of one inequality with the roots, Octogon [1] S. Math. Mag., 12(2004), no.1, 142-146. [2] J. Sándor, The Huygens inequality and its applications to convexity, Octogon Math. Mag., 10(2002), no.1, 251-254.

5

On evaluation of

k X √ n

i

i=1

√ √ √ √ Let Sn,k = n 1 + n 2 + · · · + n k. The function f (x) = n x = x1/n , x > 0, is strictly increasing, concave function µ ¶ 1 1 1 1/n−1 00 0 , f (x) = − 1 x1/n−2 ≤ 0. f (x) = x n n n Thus Z

k

f (1) + f (2) + · · · + f (k − 1)
0. By the Lagrange mean-value theorem: µ ¶ 1 1 − g(0) = g 0 (ξn ), g n n so 1

k 1+ n = k + Here 0 < ξn
0 (≥ 0), f (2k) (η) > 0 (≥ 0), we get relation (4). Corollary. If f (4) is continuous on [a, b] and f (4) > 0 (≥ 0) on (a, b), then ¶ ¶ µ µ Z b (b − a)3 00 a + b a+b + . (7) f (t)dt > (≥)(b − a)f f 2 24 2 a This is a consequence of (4) for k = 2. 2. Applications 1) For all a > 0 one has the double inequality r 2a + 2 e 1 ¶ < 1+ . 0, b = a + 1, 1 f1 (t) = , f2 (t) = − ln t. Then we get t Z a+1 Z a+1 1 2 2a + 1 dt > and ln tdt < ln , t 2a + 1 2 a a and after some elementary calculations, we get (8). Remark. There are many consequences of the double inequality (8). For example, it easily implies the limit ¶ ¸ · µ e 1 a = . lim a e − 1 + a→∞ a 2 2) For all a > 0 one has 1 2a + 2 6(2a+1) e 2 ¶ < e 0 on (a, b), and that f 00 (t) < 0 on (a, b). Then ¶ µ ¶¸ · µ (b − a)2 00 a + b a+b −1 + F . (10) θ(f ) < F F 2 24 2 Proof. Clearly F satisfies the conditions of (7), so µ ¶ µ ¶ Z b a+b (b − a)3 00 a + b F (t)dt > (b − a)F + F . 2 24 2 a By the Newton-Leibnitz formula, and after dividing with (b − a), since F is strictly increasing, we get immediately (10). For an application of (10), let us consider f (x) = ln x, 0 < a < b. Then 1 1 24 F (x) = , f 00 (x) = − 2 < 0, f (5) (x) = 5 > 0, so (10) gives x x x θ(f ) < By letting a = (11) becomes

(a + b)3 3 · 2 . 8 a + ab + b2

√ √ 3 x, b = 3 y in (11), after some elementary transformations, x−y L(x, y) = < ln x − ln y

µ√ √ ¶3 3 x+ 3y , 2

(12)

where L is the logarithmic mean of 0 < x < y. Inequality (12) is due to T.P. Lin [1]. See also [3] for related results.

References [1] T.P. Lin, The power mean and the logarithmic mean, Amer. Math. Monthly, 81(1974), 879-883. 162

[2] J. Sándor, On Hadamard’s inequality (Hungarian), Mat. Lapok (Cluj), 87(1982), 427-430. [3] J. Sándor, Some integral inequalities, Elem. Math. (Basel), 43 (1988), 177-180. [4] J. Sándor, On the theorem of finite increments (Hungarian), Mat. Lapok (Cluj), 99(1994), 361-362.

8

On certain inequalities for the number e The following inequality refines many known relations (see e.g. [3]) r 2a + 2 e 1 2a + 1 ¶a < 1 + < 0. 2a + 1 a 2a 1 1+ a

(1)

Inequality (1) appeared as an application of Hadamard’s inequality (see [1]). For refinements, with applications, see also [2]. We now prove that (1) will enable us to determine the best constants α > 0 and β > 0, such that 1+

α e β ¶n ≤ 1 + , ≤µ n n 1 1+ n

n = 1, 2, . . .

(2)

1 Remark that (1) immediately gives that the best β is β = . Indeed, if (2) 2 is true, then by (1) one must have 2n + 2 1 β =1+ 0, or equivalently µ ¶ ¶ µ ¶ µ 1 x 2x + 1 1 x ex 1+ =e . + e ln 1 + <e+ x x x+1 x+1 This may be written in a more acceptable form as ¶ µ 1 x µ ¶ 1+ 1 x 2x + 1 x + ln 1 + < . e x x+1 First, remark that, by (1) one can write µ ¶ 1 x 1+ 2x + 1 x < e 2x + 2 164

(5)

(∗)

and

µ ¶ 2x + 1 1 x 0. From (∗), (∗∗) follows relation (5). We have proved that f 0 (x) > 0, x > 0. Now this implies f (n) ≥ f (1) =

e − 1 for all n = 1, 2, . . . 2

Therefore, this is the best value of α. Since it is known that · µ ¶ ¸ 1 n e lim n e − 1 + = , n→∞ n 2 we get

1 lim f (n) = , n→∞ 2

1 so β ≥ , by reobtaining the result obtained in the first part of this note. 2 e 1 All in all, sup α = − 1, inf β = in relation (2). The equality is attained 2 2 in the left side (and it is not on the right side).

References [1] J. Sándor, On Hadamard’s inequality (Hungarian), Mat. Lapok, 89(1982), 427-430. [2] J. Sándor, Some integral inequalities, Elemente der Math. (Basel), 43(1988), 177-180. [3] J. Sándor, On certain inequalities for the number e, Octogon Math. Mag., 10(2002), no.2, 852-853.

9

The Jensen integral inequality The double inequality µ ¶ Z b a+b 1 f (a) + f (b) f ≤ f (x)dx ≤ 2 b−a a 2

(1)

which is valid for all convex functions f : [a, b] → R, is known in the literature as the Jensen-Hadamard inequalities. Particularly, the right side of (1) is called 165

the Jensen integral inequality. Recently, inequalities (1) has been extended for 2k-times (k ≥ 1) differentiable functions ([3], [1]) and various applications ([4], [5], [2]) were given. For a multivariate generalization we quote [2]. Let us now suppose that f : [a, b] → R has a strictly increasing derivative on [a, b]. By Lagrange’s mean-value theorem easily follows f (x) − f (y) < f 0 (x)(x − y) for all x, y ∈ [a, b]. By integrating with respect to x, we get Z b f (x)dx < (b − a)f (y) − y[f (b) − f (a)] + λ = g(y),

(2)

(3)

a

where

Z

b

λ=

Z xf 0 (x)dx = bf (b) − af (a) −

a

b

f (x)dx a

and g : [a, b] → R is defined as above. Clearly, g 0 (y) = (b − a)f 0 (y) − [f (b) − f (a)] so by the Lagrange mean-value theorem g 0 (y0 ) = 0 for some y0 ∈ (a, b). Since f 0 is strictly increasing, obviously g 0 (y) > g 0 (y0 ) = 0 for y > y0 and g 0 (y) < g 0 (y0 ) = 0 for y < y0 . This means that y0 is a minimum-point for the function g, that is g(y0 ) ≤ g(y) for all y ∈ [a, b]. (4) We can now formulate the following result: Theorem. Let f : [a, b] → R have a strictly increasing derivative on [a, b]. Then µ · ¸ ¶ Z b b−a f (b) − f (a) bf (b) − af (a) f (x)dx < f (y0 ) − y0 + (5) 2 b−a b−a a where y0 is defined by the equality f 0 (y0 ) =

f (b) − f (a) . b−a

(6)

For this choose of y = y0 , inequality (5) is the strongest. Proof. We apply (3) for y = y0 and then inequality (4) for the function g. We omit the details. Remark. Clearly, inequality (5) is valid for all y0 ∈ [a, b], but for y0 given by (6) we get the best result. 166

Corollary. Under the same condition we have: · µ ¶ ¸ Z b a+b f (a) + f (b) b−a b−a f + < [f (a) + f (b)]. f (x)dx < 2 2 2 2 a

(7)

a+b ∈ [a, b], by (4) and (5), after a simple compu2 tation µ we get ¶ the first relation in (7). The second inequality is a consequence a+b f (a) + f (b) of f < which is true, since by the mean-value theorem 2 2 µ ¶ µ ¶ a+b b−a 0 a+b b−a 0 f − f (a) = f (c1 ) < f (b) − f = f (c2 ), 2 2 2 2 Proof. Selecting y =

for all c1 < c2 .

1 Applications. 1) Choose f1 (x) = and f2 (x) = − log x in (5), for 0 < x √ a < b. Then y0,1 = ab = G(a, b) = G the geometric mean of a and b; b−a = L(a, b) = L, the logarithmic mean of a and b. We now y0,2 = log b − log a select f3 (x) = x log x. Denoting 1 I = I(a, b) = e

µ

bb aa

1 ¶ b−a

, a 6= b,

the so-called identric mean of a and b, we can remark that Z b 1 b log b − a log a log I(a, b) = log xdx = − 1. b−a a b−a One gets y0,3 = I(a, b). After some elementary transformations we obtain the relations G

167

bk − ak k(b − a)

1 ¶ k−1

(9)

where b > a > 0 and k > 1. In fact, (9) is valid also for k = 1 in sense that a+b > lim k→1 2

µ

bk − ak k(b − a)

1 ¶ k−1

= I(a, b),

after a little limit calculus. This is valid according to the last inequality in (8). We note that, by more difficult arguments it can be proved that (9) is valid for all real k, k 6∈ {−1, 0, 1}. 3) For f (x) = xex and using the ”exponential mean” ([6]) E(a, b) = by taking into account of Z

b

beb − aea − 1, eb − ea

a 6= b,

xex dx = (eb − ea )E(a, b)

a

we can derive from (5): ¶¶ µ µ b b2 eb − a2 ea e − ea < E(a, b) 2 + log . b−a eb − ea

(10)

References [1] H. Alzer, A note on Hadamard’s inequality, C.R. Math. Rep. Acad. Sci. Canada, 11(1989), 255-258. [2] S.S. Dragomir, J.E. Peˇcarić, J. Sándor, A note on the Jensen-Hadamard inequalities, Rev. d’Anal. Num. Th. Approx., 19(1990), 29-34. [3] J. Sándor, Some integral inequalities, Elem. Math., 43(1988), 177-180. [4] J. Sándor, Sur la fonction Gamma, Publ. C.R. Math. Neuchˆ atel, Serie I, 21(1989), 4-7. [5] J. Sándor, An application of the Jensen-Hadamard inequality, Niuew Arch Wisk., (4)8(1990), no.1, 63-66. [6] J. Sándor, Gh. Toader, On some exponential means, Seminar on Math. Analysis, Preprint no.7, 1990, 35-40. [7] K.B. Stolarky, The power and generalized logarithmic means, Amer. Math. Monthly, 87(1980), 545-548. 168

10

Generalizations of certain integral inequalities

1. Introduction Let f : [a, b] → R, a < b, be a (continuous) convex function. The double inequality µ ¶ Z b a+b 1 f (a) + f (b) f ≤ f (x)dx ≤ (1) 2 b−a a 2 which is known as the Hadamard (or Jensen-Hadamard) inequalities, has been recently intensively studied by some authors. Many applications in different branches of Mathematics have been given ([6-12]), and certain extensions, generalizations ([6], [7], [2], [1], [13], [14]) as well as refinements ([2], [3], [4]) are know. The aim of this paper is to obtain certain generalized version of the two sides of (1) for expressions of type Z b p(x)f (x)dx a = (Ep,f ) (2) Ep,f (a, b) = Z b p(x)dx a

where p is strictly positive monotone function. A key tool in proof will be the classical Chebyshev inequality ([5]): Let f1 , f2 : [a, b] → R be monotone in the same sense. Then Z b Z b Z b 1 1 1 f1 (x)f2 (x)dx ≥ f1 (x)dx · f2 (x)dx. (3) a−b a b−a a b−1 a If f1 and f2 are monotone in the opposite sense, then the sign of inequality is reversed in (3). We note that for f1 (x) ≡ 1 or f2 (x) ≡ 1 one has equality in (3). Let f be convex, p strictly positive, and f and p monotone in the same sense. Then (3) applied to f1 = f , f2 = p, and with the left side of (1) immediately give µ ¶ a+b Ep,f ≥ f . (4) 2 If f is convex and f and p are monotone in the opposite sense, by the same remark we have f (a) + f (b) Ep,f ≤ . (5) 2 We will obtain results of type (4) and (5) or refinements, for convex functions f without monotonicity properties. In what follows, p will be always a strictly positive function. 169

2. Results First prove that Theorem 1. Let f be a convex function. Then Ep,f ≥ f (A) + f+0 (A)Cp

(6)

a+b where A = and Cp = Cp (a, b). If p is increasing, then Cp ≥ 0, while for 2 decreasing p one has Cp ≤ 0. (7) Thus, when f+0 (A) ≥ 0 and p increasing, (4) holds true. If f+0 (A) ≤ 0 and p is decreasing, the same inequality (4) is valid. Proof. Since f is convex on [a, b], it is known that f (x) − f (A) ≥ f+0 (A)(x − A).

(8)

Multiplying both sides of (8) with p(x) > 0 and integrating term-by-term, we get Z b Z b Z b 0 p(x)f (x)dx − f (A) p(x)dx ≥ f+ (A) p(x)(x − A)dx. (9) a

a

Let

a

Z Kp =

b

p(x)(x − A)dx. a

We shall prove that for increasing p one has Kp ≥ 0. Indeed, since f1 (x) = p(x) and f2 (x) = x − A are increasing, by (3) Z b Z b 1 Kp ≥ p(x)dx (x − A)dx = 0, b−a a a Z b by (x − A)dx = 0. Thus Cp ≥ 0. When p decreases, an analogous proof can a

be made, giving (7). Remark. When f is convex with f+0 (A) ≥ 0, one obtains a refinement of (4), Ep,f ≥ f (A) + f+0 (A)Ep,f2 ≥ f (A), (10) where f2 (x) = x − A (see the proof of Theorem 1). If f is strictly convex with f+0 (A) > 0, then (10) holds with strict inequalities. Application. Set p(x) = exp x and denote E = E(a, b) =

beb − aea − 1, eb − ea 170

aA (12) with f (x) = x. Let f (x) = x2 in (11). By Z

b

x2 ex dx

a

Z

b

= ex dx

eb b2 − ea a2 eb − ea − 2E

a

(which gives a new exponential mean), we obtain the inequality eb b2 − ea a2 > 2E(A + 1) − A > A2 + 2E. eb − ea

(13)

We now improve relation (5). Theorem 2. Let f be convex with f (b) ≥ f (a). If p is a decreasing function, then Ep,f ≤ f (a) +

f (b) − f (a) b−a

Z

b

(x − a)p(x)dx ≤ a

f (a) + f (b) . 2

(14)

The same is valid when f (b) ≤ f (a) and p is increasing. Proof. Since f is convex on [a, b], one has f (x) ≤ f (a) +

f (b) − f (a) (x − a) = K(x) b−a

(15)

which means that f (x) ≤ K(x), where intuitively the set {(x, y) : x ∈ [a, b], y = K(x)} represents the line segment joining the points (a, f (a)), 171

(b, f (b)) of a graph of f . Multiplying both sides of (15) with p(x) > 0 and integrating, one gets µZ b ¶ Z b Z f (b) − f (a) b p(x)f (x)dx ≤ p(x)dx f (a) + (x − a)p(x)dx. (16) b−a a a a Since p(x) decreases, inequality (3) holds with reversed sign of inequality, thus Z b Z b Z b Z 1 b−a b (x − a)p(x)dx ≤ (x − a)dx p(x)dx = p(x)dx. b−a a 2 a a a By f (b) − f (a) ≥ 0 and taking into account of (16), we obtain relation (14). This holds true also when f (b)· ≤ f (a) ¸ and p is increasing. 1 Remarks. 1) The function f : − , 1 → R, f (x) = x2k (k ≥ 1 integer) µ ¶ 2 1 = 2−2k , without being monotone. Thus is convex with f (1) = 1 > f − 2 for an arbitrary strictly positive, decreasing function p one has Z 1 x2k p(x)dx ¶ Z µ 1 −2 (22k − 1)2−2k+1 1 1 −2k ≤2 + p(x)dx x+ Z 1 3 2 − 12 p(x)dx − 12

≤ (22k + 1)2−2k−2 .

(17)

With analogous proof we can say that, when f is concave, f (b) ≥ f (a), and p is increasing (or f (b) ≤ f (a) but p decreases) the inequalities in (14) are reversed. For a generalization, the positivity of f is needed: Theorem 3. Let f be positive and convex with f (b) ≥ f (a). If n ≥ 1 is a positive integer, and p is decreasing, then µ ¶ Z n µ ¶ X n k f (b) − f (a) n−k b Ep,f n ≤ f (a) (x − a)n−k p(x)dx k b−a a k=0

≤

n µ ¶ k X n f (a)(f (b) − f (a))n−k k=0

f k (a)

k

n−k+1

(f (a))k ).

.

(18)

(Here = Proof. Since f is positive, by (15) and Newton’s binomial theorem we immediately obtain the middle inequality of (18). The last inequality follows by (3) (reversed inequality) and some simple computations. 172

Remark. When p(x) ≡ 1, the condition f (b) ≥ f (a) is not necessary. (We note that for this p in Theorem 1,2 only the convexity of f is needed). In this case for n = 2 one obtains the relation Z b 1 1 f 2 (x)dx ≤ (f 2 (a) + f (a)f (b) + f 2 (b)) (19) b−a a 3 which appeared (with applications) also in [12]. A generalization of the above type can be proved for concave functions, too: Theorem 4. Let f be positive and concave and n ≥ 1 a positive integer. Then Z b n µ ¶ X n n−k 0 k Ep,f n ≤ f (A)(f+ (A)) (x − A)k p(x)dx. (20) k a k=0

If f+0 (A) ≥ 0 (where A = can be majored by

a+b ) and p is decreasing, the right side of (20) 2

n µ ¶ X n n−k 1 + (−1)k f (A)(f+0 (A))k · 2−k−1 (b − a)k . k k+1

(21)

k=0

Proof. One has (f being concave) 0 < f (x) ≤ f (A) + f+0 (A)(x − A) so by Newton’s binomial theorem one obtains (after multiplication with p(x) > 0 and integration) relation (20). For (21) remark that (3) can be used (with ”≤”). Since 1 b−a

Z

b

(x − A)k dx = 2−k−1

a

1 + (−1)k (b − a)k k+1

(easy calculation) we get the desired result. Remark. For p(x) ≡ 1, n = 2, f positive and concave one obtains the inequality Z b 1 (b − a)2 . (22) f 2 (x) ≤ f 2 (A) + (f+0 (A))2 b−a a 12

References [1] H. Alzer, A note on Hadamard’s inequalities, C.R. Math. Rep. Acad. Sci. Canada, 11(1989), 155-258. [2] S.S. Dragomir, J.E. Peˇcarić, J. Sándor, A note on the Jensen-Hadamard inequalities, Rev. d’Anal. Num. Th. Approx., 19(1990), 19-34. 173

[3] S.S. Dragomir, Two mappings in connection to Hadamard’s inequalities, J. Math. Anal. Appl., 167(1992), 49-56. [4] S.S. Dragomir, D.M. Milosević, J. Sándor, On some refinements of Hadamard’s inequalities and applications, Univ. Beograd Publ. Elekt. Fak., Ser. Mat., 4(1993), 3-10. [5] D.S. Mitrinović, Analytic inequalities, Springer Verlag, 1970. [6] J. Sándor, On Hadamard’s inequality (Hungarian), Matematikai Lapok (Cluj), 87(1982), 427-430. [7] J. Sándor, Some integral inequalities, Elem. Math., 43(1988), 177-180. [8] J. Sándor, Remark on a function which generalizes the harmonic series, C.R. Bulg. Acad. Sci., 41(1988), 19-21. [9] J. Sándor, Sur la fonction Gamma, C.R.M.P. Neuchˆ atel, Série I, Fasc. 21, 1989, 4-7. [10] J. Sándor, Inequalities for means, Proc. 3-th Symposium of Math. and Appl., 1989, Timi¸soara, 87-90. [11] J. Sándor, An application of the Jensen-Hadamard inequality, Niuew Arch. Wiskunde, (4)8(1990), 63-66. [12] J. Sándor, On the identric and logarithmic means, Aequationes Math., 40(1990), 261-270. [13] J. Sándor, On the Jensen-Hadamard inequality, Studia Univ. Babe¸sBolyai, 36(1991), 9-15. [14] J. Sándor, On certain extensions of the Jensen-Hadamard inequalities, submitted. [15] J. Sándor, Gh. Toader, On some exponential means, Seminar Math. Analysis, Babe¸s-Bolyai Univ., Preprint nr.7, 1990, 35-40. [16] Gh. Toader, An exponential mean, Seminar Math. Analysis, Babe¸sBolyai Univ., Preprint nr.5, 1988, 51-54. 174

11

The Chebyshev integral inequality let (ai ), (bi ), (pi ), i = 1, n, be sequences of real numbers, such that (ai − aj )(bi − bj ) ≥ 0,

i, j = 1, n

pi ≥ 0. Then pi pj (ai − aj )(bi − bj ) ≥ 0 for all i, j and letting

(1) X

(2) in this inequality,

i,j

after multiplication, we get: !Ã n ! ! Ã n Ã n !Ã n X X X X pi ai bi ≥ pi ai pi bi . pi i=1

i=1

i=1

(3)

i=1

This is weighted Chebyshev inequality for sequences. An example for (1) is when (ai ) and (bi ) are syncrone sequences (i.e. having the same type of monotonity). Let (1) be replaced by (ai − aj )(bi − bj ) ≤ 0,

i, j = 1, n.

In this case one has, by a similar proof ! !Ã n ! Ã n Ã n !Ã n X X X X pi bi . pi ai pi ai bi ≤ pi i=1

i=1

i=1

(10 )

(30 )

i=1

An example of (10 ) is when (ai ), (bi ) are asyncrone sequences, i.e. when they have different type of monotonity. Let now p, f, g : [a, b] → R be real variable functions, such that p ≥ 0 is integrable and f, g are monotone on [a, b]. Then it is well-known that f, g are integrable, too. Let ∆ = {a = x0 < · · · < xi−1 < xi < · · · < xn = b} be an arbitrary division of [a, b] and let σ∆ (f, ξ) =

n X

f (ξi )(xi − xi−1 )

i=1

be the Riemann-sum associated to f , where ξ = (ξ1 , . . . , ξn ) and ξi ∈ [xi−1 , xi ] are arbitrary. Now, let f, g be syncrone functions on [a, b] (i.e. having the same type of monotonity) and put ai = f (ξi ), bi = g(ξi ), pi = p(ξi )(xi − xi−1 ) in (3). One gets: n n X X p(ξi )(xi − xi−1 ) p(ξi )g(ξi )(xi − xi−1 ) ≥ i=1

i=1

175

≥

n X

n X

p(ξi )f (ξi )(xi − xi−1 )

i=1

p(ξi )g(ξi )(xi − xi−1 ).

(4)

i=1

Now, letting k∆k = sup{xi −xi−1 } → 0 in (4), since the occurring Riemann sums have a limit equal to the integral of the corresponding function, we get: Z

Z

b

Z

b

p(x)dx a

Z

b

p(x)f (x)g(x)dx ≥

b

p(x)f (x)dx

a

a

p(x)g(x)dx.

(5)

a

This is Chebyshev’s integral inequality with weights for syncrone functions. When the functions f, g are asyncrone on [a, b], we get from (30 ): Z

Z

b

a

Z

b

p(x)dx a

Z

b

p(x)f (x)g(x)dx ≤

b

p(x)f (x)dx a

p(x)g(x)dx

(6)

a

the Chebyshev inequality for asyncrone sequences. Inequalities of type (3), (4) are very important in many parts of Mathematics ([1]). For applications in the Theory of Means and Number Theory, we quote [2], [3], [4], [5], [6]. Let now f, g be syncrone functions, and consider the function defined by h(t) = g(a + b − t), t ∈ [a, b]. Then clearly f, h are asyncrone functions, and applying (6) for (p, f, g) := (p, f, h) one gets Z

Z

b

p(x)dx a

b

p(x)f (x)g(a + b − x)dx ≤ a

Z

Z

b

≤

p(x)f (x)dx a

b

p(x)g(a + b − x)dx.

(7)

a

Let us now suppose that p has the following property: p(a + b − x) = p(x),

x ∈ [a, b].

(8)

Then, by the substitution a + b − x = t one gets Z

Z

b

b

p(x)g(a + b − x)dx =

p(t)g(t)dt,

a

a

and applying (5), by taking into account of (7), we get the following result: Theorem. Let f, g : [a, b] → R be two syncrone functions and let p : [a, b] → R be a nonnegative integrable function such that (8) is true. Then Z

Z

b

p(x)dx a

Z

b

p(x)f (x)g(a + b − x)dx ≤ a

p(x)f (x)dx a

176

Z

b

b

p(x)g(x)dx a

Z

Z

b

≤

b

p(x)dx a

p(x)f (x)g(x)dx.

(9)

a

Remark. When p(x) ≡ 1, (8) is true. But let p(x) = cos x, [a, b] = ¸ −π π . Then by cos(−x) = cos x, (8) is again true. Here cos x ≥ 0 on , · 2 2¸ −π π , . These two cases, give the double-inequalities 2 2 ·

Z (b − a)

Z

b

f (x)g(a + b − x)dx ≤ a

Z

b

g(x)dx

a

Z

b

f (x)dx a

b

≤ (b − a)

f (x)g(x)dx

(10)

a

Z 2

π 2

− π2

Z cos xf (x)g(−x)dx ≤ Z ≤2

π 2

− π2

π 2

− π2

Z cos xf (x)g(x)dx

π 2

− π2

cos xg(x)dx

cos xf (x)g(x)dx

(11)

where, in both cases f, g are syncrone functions.

References [1] G.H. Hardy, J.E. Littlewood, G. Pólya, Inequalities, Oxford, 1934. [2] J. Sándor, On Jordan’s arithmetical function. Math. Student, 52(1984), 91-96. [3] J. Sándor, On certain integral inequalities, Octogon Math. Mag., 5 (1997), 29-34. [4] J. Sándor, On means generated by derivatives of functions, Int. J. Math. E. Sci. Tech., 28(1997), 146-148. [5] J. Sándor, Gh. Toader, Some general means, Czechoslovak Math. J., 49(124)(1999), 53-62. [6] J. Sándor, Inequalities for generalized convex functions with applications, II, Studia Univ. Babe¸s-Bolyai Math., XLVI, 2001, 79-92. 177

12

On Fink’s inequality The following interesting inequality is due to A.M. Fink [1]: Theorem 1. If f > 0 and log f is convex on R, then Z

1

f (x + vt) cos −1

πt 2 dt ≤ (f (x + v) + f (x − v)), 2 π

x, v ∈ R.

(1)

The aim of this note is to prove that relation holds true if f is convex on R. Lemma. If g : I → R is a strictly positive, log-convex function, then it is convex in I (real interval). Proof. If g : I → R (I ⊂ R, interval) is a strictly positive, log-convex function, then log g(λa + (1 − λ)b) ≤ λ log g(a) + (1 − λ) log g(b) for all λ ∈ [0, 1]; a, b ∈ I, implying g(λa + (1 − λ)b) ≤ (g(a))λ (g(b))1−λ . By Hölder’s inequality (see e.g. [2]) one has (g(a))λ (g(b))1−λ ≤ λg(a) + (1 − λ)g(b), since λ + (1 − λ) = 1, λ ≥ 0. Thus, g is convex. To prove (1) for convex f , first note that Z

1

πt I= f (x + vt) cos dt = 2 −1

Z

1

(f (x + vt) + f (x − vt)) cos 0

πt dt. 2

(2)

Put gx,v (t) = f (x + vt) + f (x − vt),

t ∈ [0, 1].

(3)

Now, since µ x ± vt =

1+t 2

¶

µ (x ± v) +

1−t 2

¶ (x ∓ v),

by convexity of f one can write gx,v (t) ≤

1+t 1−t 1+t 1−t f (x + v) + f (x − v) + f (x − v) + f (x + v) 2 2 2 2 178

= f (x + v) + f (x − v),

t ∈ [0, 1].

So by (2) we have Z I ≤ (f (x + v) + f (x − v))

1

cos 0

πt 2 dt = (f (x + v) + f (x − v)). 2 π

Equality occurs only when gx,v is linear. Then from (3) it follows that f must be a constant. The given proof shows that the following generalization of (1) is valid. Theorem 2. If f is convex function on R, and c is a nonnegative, even function on [−1, 1], then Z 1 Z f (x + v) + f (x − v) 1 f (x + vt)c(t)dt ≤ c(t)dt, x, v ∈ R. 2 −1 −1

References [1] A.M. Fink, Two inequalities, Univ. Beograd Publ. Elektrotehn. Fak., Ser. Mat., 6(1995), 48-49. [2] D.S. Mitrinović, Analytic inequalities, Springer Verlag, 1970.

13

An extension of Ky Fan’s inequalities

Let xk , k = 1, n, be positive real numbers. The arithmetic respectively geometric means of xk are x1 + · · · + xn , n √ G = G(x1 , . . . , xn ) = n x1 . . . xn .

A = A(x1 , . . . , xn ) =

Let f : I → R (I interval), and suppose that xk ∈ (a, b). Define the functional arithmetic, respectively geometric ”means” by Af = Af (x1 , . . . , xn ) =

f (x1 ) + · · · + f (xn ) n

and Gf = Gf (x1 , . . . , xn ) =

p n

f (x1 ) . . . f (xn ).

Clearly, Af and Gf are means in the usual sense, if min{x1 , . . . , xn } ≤ Af ≤ max{x1 , . . . , xn } and min{x1 , . . . , xn } ≤ Gf ≤ max{x1 , . . . , xn }. For 179

example, when I = (0, +∞) and f (x) = x, Af ≡ A, Gf ≡ G; when I = (0, 1) and f (x) = 1 − x, Af = A0 , Gf = G0 , are indeed means in the above sense. The following famous relations are well-known: G ≤ A for xk > 0, k = 1, n

(1)

µ ¸ 1 0, . 2

(2)

A G ≤ 0 for xk ∈ 0 G A

The first is the arithmetic-geometric inequality, while the second is the Ky-Fan inequality (see e.g. [2], [3], [4]). Now, even if Af and Gf are not means in the usual sense, the following extension of (2) may be true: G A ≤ . Gf Af

(3)

This inequality (with other notations) is stated in OQ.633, in [1]. We now prove (3) for certain particular f . Theorem. Let id : R → R, id(x) = x and suppose that f : I → R satisfies id the following conditions: f and ln are concave functions. Then inequality f (3) holds true. id Proof. By concavity of ln one can write: f x1 + · · · + xn · ¸ x1 xn 1 n ¶ ≥ ln ln µ + · · · + ln , x1 + · · · + xn f (x1 ) f (xn ) n f n i.e. ln Therefore

A G ≥ ln . f (A) Gf G A ≤ . Gf f (A)

(4)

Now, since f is concave, one has f (x1 ) + · · · + f (xn ) Af = ≤f n and by (4) this gives (3). 180

µ

x1 + · · · + xn n

¶ = f (A),

µ ¸ 1 x Remark 1. Let I = 0, and f (x) = 1 − x, then g(x) = ln has a 2 1−x derivative 1 1 g 0 (x) = + , x 1−x so g 00 (x) = −

1 x2 − (1 − x)2 2x − 1 1 + = = 2 ≤ 0. 2 2 2 2 x (1 − x) x (1 − x) x (1 − x)2

id are concave functions, and (4) gives Ky Fan’s inf 1 equality (2). One has equality for xk = , k = 1, n. 2 id Remark 2. There are many functions f : I → R such that f and ln f are simultaneously concave. Put e.g. f (x) = ln x. Then Therefore f and ln

g(x) = ln

x = ln x − ln ln x. ln x

One has g 00 (x) =

− ln2 x + ln x + 1 ≤0 x2 ln2 x

√ √ 1+ 5 1+ 5 2 if ln x ≥ , i.e. x ≥ e = x0 . (Take I = [x0 , +∞)). 2 Remark 3. Without concavity of f , holds true (4).

References [1] M. Bencze, OQ.633, Octogon Math. Mag., 9(2001), no.1, 679-680. [2] J. Sándor, On an inequality of Ky Fan, III, Intern. J. Math. Ed. Sci. Tech., 32(2001), no.1, 133-160. [3] J. Sándor, T. Trif, A new refinement of the Ky Fan inequality, Math. Ineq. Appl., 2(1999), no.4, 529-533. [4] E. Neumann, J. Sándor, On the Ky Fan inequality and related inequalities, I, Math. Ineq. Appl., 5(2002), 49-56; for II, see Bull. Austral. Math. Soc. (to appear). 181

14

A converse of Ky Fan’s inequality

µ ¸ 1 Let xi ∈ 0, , i = 1, n, and let An , Gn , Hn denote the arithmetic, geo2 metric, resp. harmonic means of these numbers. Put A0n , G0n , Hn0 for the corresponding means of the numbers 1 − xi . The famous inequality of Ky Fan (see [1]) states that An Gn ≤ 0 . (1) 0 Gn An ¸ · 1 Suppose that m > 0 and xi ∈ m, . Then the following converse of (1) 2 is true: Theorem. ¸ · An Gn 1 . (2) ≤ 0 exp (An − Gn ) A0n Gn m(1 − m) Proof. We shall obtain a slightly stronger relation. Let us define ½³ ¾ · ¸ x´ 1 1 x exp 1− , where x ∈ m, . f (x) = 1−x m 1−m 2 Then

f 0 (x) 1 1 = − ≤ 0 for x ≥ m, f (x) x(1 − x) m(1 − m) · ¸ 1 since the function g(x) = x(1 − x) is strictly increasing on 0, . Thus the 2 ¸ · 1 function f is non-increasing on m, . Suppose that m ≤ x1 ≤ x2 ≤ · · · ≤ 2 1 xn ≤ . Then, since m ≤ Gn ≤ An , we have f (An ) ≤ f (Gn ) so that 2 · ¸ An 1 Gn exp (An − Gn ) . (3) ≤ A0n 1 − Gn m(1 − m) This is slightly strongerpthan (2), since 1 − Gn ≥ G0n . This follows by √ the well-known inequality n (1 + a1 ) . . . (1 + an ) ≥ 1 + n a1 . . . an (see e.g. ui √ [2], Theorem 64, p.61), which for ai = , ui , vi > 0, implies n u1 . . . un + v i p √ n n v ...v (u1 + v1 ) . . . (un + vn ). Apply then this inequality to ui = xi , 1 n ≤ vi = 1 − xi in order to get Gn + G0n ≤ 1, xi ∈ (0, 1). Thus (2) follows. 182

References [1] E.F. Beckenbach, R. Bellman, Inequalities, Springer Verlag, 1961. [2] G.H. Hardy, J.E. Littlewood, G. Pólya, Ineqialities, Cambridge Univ. Press, Second Reprinted ed., 1964.

15

A refinement of Gn + G0n ≤ 1

Suppose that xi ∈ (0, 1), i = 1, 2, . . . , n, and let Gn = Gn (xi ) denote the √ geometric mean of xi , i.e. Gn = n x1 . . . xn . Put p G0n = n (1 − x1 ) . . . (1 − xn ). The well-known inequality p √ √ n u1 . . . un + n v1 . . . vn ≤ n (u1 + v1 ) . . . (un + vn ),

ui , vi ≥ 0

applied to ui = xi , vi = 1 − xi implies the relation Gn + G0n ≤ 1 (see also [2]). By using the optimization theory of concave functions, in what follows we shall prove the following refinement: Theorem. (Gn + G0n )n ≤ (Gn−1 + G0n−1 )n−1 ≤ 1 for all n ≥ 2. Proof. Let us consider the application f : (0, 1) → R, defined by f (x) = (x1 . . . xn−1 x)1/n + [(1 − x1 ) . . . (1 − xn−1 )(1 − x)]1/n . We have 1 1 1 1 1 1 (x1 . . . xn−1 ) x x n −1 − [(1 − x1 ) . . . (1 − xn−1 )] n (1 − x) n −1 , n n Gn−1 so f 0 (x) = 0 iff x = x0 = . Gn−1 + G0n−1 Since µ ¶ 1 1 1 1 00 f (x) = − 1 (x1 . . . xn−1 ) n x n −2 n n µ ¶ 1 1 1 1 + − 1 [(1 − x1 ) . . . (1 − xn−1 )] n (1 − x) n −2 ≤ 0 n n

f 0 (x) =

we observe that f is a concave function. It is well known (see e.g. [1]) that then x0 must be a maximum point on (0, 1), implying f (xn ) ≤ f (x0 ). After some simple calculations this gives Gn + G0n ≤ (Gn−1 + G0n−1 ) i.e. the first relation of the Theorem. 183

n−1 n

,

References [1] A.W. Roberts, D.E. Varberg, Convex functions, Academic Press, New York and London, 1973. [2] J. Sándor, On an inequality of Ky Fan, Babe¸s-Bolyai Univ. Preprint nr.7, 29-34.

16

On Alzer’s inequality H. Alzer [1] discovered the following result: Theorem. If r is a positive real number and n is a positive integer, then n ≤ n+1

Ã (n + 1)

n X

r

i /n

i=1

n+1 X

!1/r r

i

.

(1)

i=1

We will obtain a proof based on mathematical induction and Cauchy’s mean value theorem of differential calculus. More precisely, we will prove that (1) holds with strict inequality. Proof. Let us denote Sr (n) = 1r + 2r + · · · + nr ,

n = 1, 2, . . . , r > 0.

Then, since Sr (n + 1) = Sr (n) + (n + 1)r , it is immediate that inequality (1) is equivalent to Sr (n) ≥ nr+1 (n + 1)r /((n + 1)r+1 − nr+1 ).

(2)

We shall deduce this inequality via mathematical induction. For n = 1 one has 1 ≥ 2r /(2r+1 − 1) which is true by 2r ≥ 1, r > 0. Now, accepting (2) for n, we try to obtain it for n + 1. By Sr (n + 1) = Sr (n) + (n + 1)r , it is easy to see that the induction step (after dividing by (n + 1)r ) can be written as µ ¶ n+2 r (n + 2)r+1 − (n + 1)r+1 ≥ . (3) (n + 1)r+1 − nr+1 n+1 Let us consider the functions f, g : [n, n + 1] → R defined by f (x) = (x + 1)r+1 ,

g(x) = xr+1 ,

x ∈ [n, n + 1].

By Cauchy’s mean-value theorem one has [f (n + 1) − f (n)]/[g(n + 1) − g(n)] = f 0 (ξ)/g 0 (ξ), 184

where ξ ∈ (n, n + 1), so (n + 2)r+1 − (n + 1)r+1 (ξ + 1)r = = (n + 1)r+1 − nr+1 ξr

µ ¶ 1 r 1+ . ξ

(4)

By 1/ξ > 1/(n + 1) in (4), we have µ r

(1 + 1/ξ) > (1 + 1/(n + 1)) =

n+2 n+1

¶r ,

implying relation (3). This finishes the proof of (2), and thus of the theorem, with strict inequality (see also [2]).

References [1] H. Alzer, On an inequality of H. Minc and L. Sathre, J. Math. Anal. Appl., 179(1993), 396-402. [2] J. Sándor, On an inequality for Alzer, J. Math. Anal. Appl., 192(1995), 1034-1035.

185

186

Chapter 5

Euler gamma function ”... Mathematics compares the most diverse phenomena and discovers the secret analogies that unite them.” (Joseph Fourier)

”... Euler’s integral appears everywhere and is inextricably bound to a host of special functions. Its frequency and simplicity make it fundamental.” (Ph. J. Davis)

187

1

A limit involving the Gamma function and the Lalescu sequence Let Lk =

p

k+1

(k + 1)! −

√ k k!. We will compute n

1X (Γ(1 + Lk ))1/Lk , n→∞ n lim

k=1

where Γ is the Euler gamma function. Put xn = (Γ(1 + Ln ))1/Ln . It is well1 known that Ln → as n → ∞ (Traian Lalescu sequence). Since the Γ-function e is continuous in (0, +∞), it follows that µ µ ¶¶e · µ ¶¸e 1 1 1 xn → Γ 1 + Γ = . e e e Now the following theorem is well-known: x1 + x2 + · · · + xn → a. Therefore the proposed If xn → a, n → ∞, then n · µ ¶¸e 1 1 limit has a value Γ . e e

2

On a sequence containing the Gamma function Let 0 < x1 < x2 < · · · < xn < . . . , and (yn ) be defined by yn =

n X

Ψ(xk ) − ln Γ(xn )

(1)

k=1

where Ψ(x) =

Γ0 (x) , with Γ and Ψ the gamma, respectively digamma funcΓ(x)

tions. Generally, the sequence (yn ) is not convergent. Indeed, suppose that (xn ) is convergent, having a limit lim xn = x0 6= a, where a ∈ (1, 2) is the single n→∞

positive root of Γ0 (x) = 0. Then (yn ) is divergent. Indeed, yn+1 −yn = Ψ(xn+1 )−[ln Γ(xn+1 )−ln Γ(xn )] and since Γ, Ψ are continuous; if (yn ) could be convergent, we would obtain: 0 = Ψ(x0 ) − [ln Γ(x0 ) − ln Γ(x0 )] = Ψ(x0 ) so Γ0 (x0 ) = 0, contradiction with the made assumption. Let us now suppose that 0 < xn+1 − xn ≤ 1. Then yn+1 − yn = Ψ(xn+1 ) − [ln Γ(xn+1 ) − ln Γ(xn )]. 188

Here ln Γ(xn+1 ) − ln Γ(xn ) = (ln Γ0 )(ξ)(xn+1 − xn ) = Ψ(ξ)(xn+1 − xn ), by the Lagrange mean-value theorem. Since it is well known that ln Γ is a convex function on (0, +∞), the digamma function Ψ will be strictly increasing, so ln Γ(xn+1 ) − ln Γ(xn ) < Ψ(xn+1 )(xn+1 − xn ) ≤ Ψ(xn+1 ), by the accepted assumption. This implies yn+1 − yn > 0, i.e. (yn ) is a strictly increasing sequence. Thus yn > yn−1 > · · · > y1 , and (yn ) will be bounded below. In any case, (yn ) will have a limit, but this limit may be +∞, if it is not bounded above. By supposing that Ψ(x1 ) + · · · + Ψ(xn ) − ln Γ(xn ) < K (K constant), for all n ≥ n0 , the sequence (yn ) will be convergent.

3

The product of consecutive factorials We will evaluate

n Y

k!, or in fact

k=1

Ã

!

n Y

k!

=

n X

ln k! =

ln Γ(k + 1).

k=1

k=1

k=1

n X

Now, the function g(x) = ln Γ(x) is known to be strictly convex and strictly increasing for x ≥ 2. By area considerations Z g(3) + · · · + g(k)
0.

(9)

By applying Euler’s formula (2) one has k kx g(x) =

k−1 Y

m!mx+i/k ¶µ ¶ µ ¶ lim µ m→∞ 1 i i i=0 x+ x + + 1 ... x + + m k k k kx 1 · 2 · 3 . . . (mk)(mk) k lim m→∞ (kx)(kx + 1) . . . (kx + mk) k−1

(m!)mkx+ 2 k mk m→∞ (km)!(km)kx

= lim

which is µ independent of x. So g is a constant (of x); implying for example ¶ 1 . By the definition of g (see (9)) one can write g(x) = g k µ ¶ µ ¶ µ ¶ 1 2 k−1 µ ¶ Γ Γ ...Γ Γ(1) 1 (2π)(k−1)/2 k k k √ g = = , k Γ(1) π by (8). Therefore, we have obtained that: µ ¶ µ ¶ 1 k kx− 2 1 k−1 Γ(kx) = Γ(x)Γ x + ...Γ x + , k k (2π)(k−1)/2

x > 0.

(10)

Particularly, for x = n, for Fk (n) in (6) we get relation (7).

5

On a limit for the quotients of Gamma functions Z Let Γ(x) =

∞

e−t tx−1 dt, x > 0 be the Euler gamma function. The follow-

0

ing asymptotic result is known as the Stirling formula for the gamma function ([1]) µ ¶¸ ³ x ´x · √ 1 1 Γ(x + 1) = 2πx 1+ +O , x > 0. (1) e 12x x2 192

1. Letting x → ax and x → bx (x > 0, a, b > 0) in (1) one obtains successively µ ¶¸ 1 ³ ax á · x 1 1 1 1 1+ [Γ(ax + 1)] x = (2πax) x +O (2) 2 e 12ax x µ 1 x

[Γ(bx + 1)] = (2πbx)

1 x

bx e

¶b ·

1 1+ +O 12bax

µ

1 x2

¶¸ 1

x

(3)

(for fixed a, b > 0). From (2) and (3) one obtains ·

Γ(bx + 1) Γ(ax + 1)

Clearly

¸1

x

µ ¶¸ 1 · x 1 1 µ ¶ 1 µ ¶b ³ ´ 1 + + O b x b e a b−a 12bx x2 = x µ · ¶¸ 1 . a e a x 1 1 +O 1+ 2 12ax x ·

1 1+ +O 12ax

µ

1 x2

(4)

¶¸ 1

x

→1

µ ¶1 b x → 1. Therefore, from (4) one can deduce the following as x → ∞ and a limit: Theorem. ¸1 · bb 1 Γ(bx + 1) x = a ea−b . (5) lim b−a x→∞ x Γ(ax + 1) a Certain particular cases of (5) are of interest. Put b = a + 1. Then one has: · ¸1 1 Γ((a + 1)x + 1) x 1 (a + 1)a+1 lim = . x→∞ x Γ(ax + 1) e aa We note that for a = 1 this gives · ¸1 1 Γ(2x + 1) x 4 lim = x→∞ x Γ(x + 1) e

(6)

(7)

which solves the OQ.1055 (see [2]). Let x = n ∈ N in (7). Since Γ(n + 1) = n!, Γ(2n + 1) = (2n)!, and (2n)! = (n + 1)(n + 2) . . . 2n, from (7) one can deduce the following limit: n! 1p 4 n lim (n + 1)(n + 2) . . . (n + n) = . (8) n→∞ n e 193

Another solution for (8) is based on Riemann’s integral. Indeed, remark that: r · µ ¶ ¸ ³ 1 1 n´ n (n + 1)(n + 2) . . . (n + n) ln = ln 1 + + · · · + ln 1 + , nn n n n so r Z 1 n (n + 1) . . . (n + n) = ln(1 + x)dx = ln 4 − 1, lim ln n→∞ nn 0 which implies relation (8). 2. For another proof of relation (7) remark that by the known formula µ ¶ √ 1 2x−1 2 Γ(x)Γ x + = πΓ(2x) 2 (see [3]) and Γ(x + 1) = xΓ(x), Γ(2x + 1) = 2xΓ(2x), one has: ¶ µ Γ(2x + 1) 22x 1 = √ Γ x+ . Γ(x + 1) 2 π Now, 1 x

µ

Γ(2x + 1) Γ(x + 1)

¶1

x

(9)

· µ ¶¸ 1 4 1 1 x = √ 1 Γ x+ . 2 ( π) x x

We shall prove that: · µ ¶¸ 1 1 x 1 1 Γ x+ = . lim x→∞ x 2 e We shall use the following asymptotic formula for Γ(x): µ ¶ µ ¶ 1 1 1 ln Γ(x) = x − ln x − x + ln 2π + O . 2 2 x

(10)

(11)

1 Putting x → x + from (11) we get 2 µ ¶ µ ¶ µ ¶ 1 1 1 1 1 Γ x+ = x ln x + − x − + ln 2π + O 2 2 2 2 x so

µ ¶ µ ¶ µ ¶ 1 1 1 1 ln Γ x + = ln x + −1+O , x 2 2 x

i.e.

· lim

x→∞

µ ¶ ¸ 1 1 ln Γ x + − ln x = −1 x 2

giving relation (10). 194

(12)

References [1] E.T. Whittaker, G.N. Watson, A course of modern analysis, Cambridge Univ. Press, 1952. [2] D.M. B˘atinet¸u-Giurgiu, M. Bencze, OQ.1055, Octogon Math. Mag., 10(2002), no.2, 1060. [3] J. Sándor, On certain expression involving Euler’s gamma function, Octogon Math. Mag., 8(2000), no.2, 505-507.

6

A limit for the Euler-Beta function Z

1

Let B(x, y) =

tx−1 (1 − t)y−1 dt, x, y > 0 be the Euler-Beta function.

0

Let a, b > 0, c, d ≥ 0 and consider the limit ([1]) 1

lim [B(ax + c, bx + d)] x =?

x→∞

(1)

The following identity, connecting the Γ and B functions, is well-known ([2]) Γ(x)Γ(y) B(x, y) = , x, y > 0. (2) Γ(x + y) Our aim is to prove the following result: Theorem. For all a, b > 0, c, d ≥ 0 one has 1

lim [B(ax + c, bx + d)] x =

x→∞

aa bb . (a + b)a+b

(3)

Proof. Since µ µ ¶¶ √ 1 1 x −x Γ(x + 1) = 2πxx e 1+ +O , 12x x2 and Γ(x + 1) = xΓ(x), we get ([2]) µ µ ¶¶ √ 1 1 x−1 −x Γ(x) = 2πxx e 1+ +O . 12x x2 By relations (2) and (3) one can write B(ax + c, bx + d) =

Γ(ax + c)Γ(bx + d) Γ((a + b)x + c + d)

195

(4)

p =

p 2π(ax+c)(ax + c)ax+c−1 e−(ax+c) 2π(bx + d)(bx + d)bx+d−1 e−(bx+d) p · 2π[(a + b)x + c + d][(a + b)x + c + d](a+b)x+c+d−1 e−((a+b)x+c+d) µ µ ¶¶ µ µ ¶¶ 1 1 1 1 1+ 1+ +O +O 2 12(ax + c) x 12(bx + d) x2 µ ¶ · . (5) 1 1 1+ +O 12[(a + b)x + c + d] x2

In what follows remark that e−((a+b)x+c+d) = e−(ax+c) e−(bc+d) , and that the following limit is true: 1

lim (Ax + B) x = 1,

A > 0, B ≥ 0.

x→∞

(6)

Now, after simplifications in (5), and by taking into account limits of type (6), and the obvious relation µ ¶¶ 1 µ x 1 1 +O →1 1+ 12(Ax + B) x2 as x → ∞, all is reduced to the calculation of a limit lim

x→∞

(ax + c)

ax+c−1 x

(bx + d)

[(a + b)x + c + d]

bx+d−1 x

(a+b)x+c+d−1 x

,

which by using again (6) and (ax + c)a (bx + d)b aa bb = , x→∞ [(a + b)x + c + d]a+b (a + b)a+b lim

gives the proof of (3). For a = 2, b = 3 and c, d ≥ 0 arbitrarily one obtains 1

lim (B(2x + c, 3x + d)) x =

x→∞

22 · 33 108 = 5 5 3125

which contains, as a particular case the known limit p 108 lim n B(2n, 3n) = . n→∞ 3125

(7)

(8)

References [1] D.M. B˘atinet¸u-Giurgiu, M. Bencze, OQ.991, Octogon Math. Mag., 10(2002), no.2, 1045. [2] E.T. Whittacker, G.N. Watson, A course of modern analysis, Cambridge Univ. Press, 1952. 196

7

On convex functions involving Euler’s Gamma function Z

∞

Let Γ(x) =

e−t tx−1 dt, x > 0 be the Euler-gamma function, for x > 0.

0

Let

f (x + 1) g(x) = , f (x) µ ¶ Ψ(x + 1) 1 h(x) = − ln Γ(x + 1), x x2 1

f (x) = (Γ(x + 1)) x ,

Γ0 (x) is the Euler digamma function. Furthermore, put K(x) = Γ(x) Ψ0 (x). In our papers [1], [2] the following results have been proved: f (x) is strictly decreasing for x > 1, but (f (x))2 /x is a. The function x strictly increasing for x ≥ 6. b. The function g(x) is strictly decreasing and strictly convex for x ≥ 6. c. The function xg(x) is strictly increasing for x > 1 and strictly concave for x ≥ 5. d. The function f (x) is strictly concave for x > 7. e. The function h(x) is strictly decreasing and strictly convex for x ≥ 6. 1 f. The function f1 (x) = (Γ(x)) x is strictly convex for x > 0, and logarithmic-concave for x ≥ 6. f (x) 1 g. & , x → ∞. x e 1 h. 0 < h(x) < for x > 0. x In these papers there are included also many other results on the above functions. Among others, in [1] is it proved that: where Ψ(x) =

K(x + 1)
0.

(1)

x2 . Then the Open Problem OQ.325 by D.M. B˘atinet¸uf (x) Giurgiu asks for the convexity of d. Let us remark that: Let d(x) =

f 0 (x) = f (x)h(x), so a simple computation gives: d0 (x) =

2x − h(x)x2 , f (x)

d00 (x) = 197

h2 x2 − 4hx − h0 x2 + 2 f

(2)

where h = h(x), f = f (x), h0 = h0 (x). Since h0 (x) =

2 Γ0 K(x + 1) 2 K 2h ln Γ − · + = − 3 2 x x Γ x x x

(where K = K(x + 1)); we get h0 x2 = x(K − 2h). This gives the following formula: h2 x2 − 4hx − h0 x2 + 2 = h2 x2 − 2hx + 2 − xK = (hx − 1)2 + 1 − xK.

(3)

Now, by relation (1) one has 1 − xK > 0, so thus, clearly, by (3) we get d00 (x > 0) (see (2)), finishing the proof of convexity of d. This solves also problem OQ.323 on the monotonicity of sequence Bn =

(n + 1)2 n2 p √ . − n n+1 n! (n + 1)!

Remark. Our results from [1], [2] have been applied in many directions (Harmonic series, Diophantine equations, Theory of inequalities, Number theory, etc.).

References atel, Série I, [1] J. Sándor, Sur la fonction Gamma, Publ. C.R.M.P. Neuchˆ 21, 1989, 4-7. [2] J. Sándor, On the Gamma function II, Publ. C.R.M.P. Neuchˆ atel, Série I, 28, 1997, 10-12. [3] D.M. B˘atinet¸u-Giurgiu, Open Problem OQ.325, Octogon Math. Mag., 8(2000), no.1, 269. [4] D.M. B˘atinet¸u-Giurgiu, Open Problem OQ.323, Octogon Math. Mag., 8(2000), no.1, 269.

8

A convexity result on (f (x))1/x

Theorem. Let f : I → R be a strictly positive, twofold differentiable function and let K(x) = (log f (x))00 , x > 0. Let us suppose that: K(x) ≥

1 , x

x>a≥0

holds true. Then the function F (x) = (f (x))1/x is convex for x > a ≥ 0. 198

Proof. After elementary computations we get: 2

x4 f 2 (x)F 00 (x) = f 2 log2 f − 2xf f 0 log f + x2 f 0 + 2xf 2 log f 2

−2x2 f f 0 + (f f 00 − f 0 )x3 ,

(5)

where f = f (x), etc. Remark that: f (x)f 00 (x) − (f 0 (x))2 = f 2 (x)K(x), so by K(x) ≥

1 , we have x 2

(f f 00 − f 0 )x3 ≥ f 2 x2 . Thus, we get from (5): 2

x4 f 2 (x)F 00 (x) ≥ f 2 log2 f − 2xf f 0 log f + x2 f 0 + 2xf 2 log f −2x2 f f 0 + f 2 x2 = (f log x + xf − xf 0 )2 ≥ 0. 1 Therefore F 00 (x) ≥ 0 (or > 0, if K(x) > , and then F is strictly convex), x thus F is convex. 1 Remarks. Condition K(x) ≥ can be written equivalently as x ((log f (x))0 − log x)0 ≥ 0, or that the function G(x) =

f 0 (x) − log x, f (x)

x>a≥0

is an increasing function. Therefore, it is not difficult to give examples. Put 1−α e.g. f (x) = xα (α < 1). Then G0 (x) = > 0. x2 Another example is f (x) = Γ(x), where Γ is Euler’s gamma function. Then 1 it is known that K(x) > for all x > 0 (see [2]), thus we obtain the convexity x of (Γ(x))1/x for x > 0. This, along with other results, appears in [2].

References [1] D.S. Mitronović, Analytic inequalities, Springer Verlag, 1970. atel, Série I, [2] J. Sándor, Sur la fonction gamma, Publ. C.R.M.P. Neuchˆ 21(1989), 4-7. 199

9

On a subadditive property of the Gamma function

1. Let a > 0, and f : [0, a] → R be a convex function defined on the interval [0, a]. In 1932 M. Petrović [1], proved the following result: If xi ∈ [0, a], i = 1, n, and x1 + · · · + xn ∈ [0, a], then f (x1 ) + · · · + f (xn ) ≤ f (x1 + · · · + xn ) + (n − 1)f (0). (1) Z ∞ Let x > 0. Then Γ(x) = tx−1 e−t dt (Euler’s Gamma Function) satisfies 0

the following properties: 1) Γ(x + 1) = Γ(x); 2) Γ(1) = 1; 3) ln Γ is a convex function. We note that Bohr and Mollerup discovered the surprising fact that 1)-3) characterize the Γ function for x > 0 (see [2]). Now let f (x) = ln Γ(x + 1), x ≥ 0. Clearly f is convex, so by applying Petrović’s inequality (1) one can deduce Γ(x1 + x2 + · · · + xn + 1) ≥ Γ(x1 + 1) . . . Γ(xn + 1)

(2)

(since f (0) = ln Γ(1) = 0) for any xi ∈ [0, a] such that x1 +x2 +· · ·+xn ∈ [0, a]. Particularly, for n = 2 one gets Γ(x1 + x2 + 1) ≥ Γ(x1 + 1)Γ(x2 + 1)

(3)

or (x1 + x2 )Γ(x1 + x2 ) ≥ (x1 x2 )Γ(x1 )Γ(x2 ), giving Γ(x1 + x2 ) ≥ if x1 , x2 ∈ (0, a], x1 + x2 ∈ (0, a]. Let (x1 − 1)(x2 − 1) ≥ 1. Then

x1 x2 Γ(x1 )Γ(x2 ), x1 + x2

(4)

x1 x2 ≥ 1, so (4) gives x1 + x2

Γ(x1 + x2 ) ≥ Γ(x1 )Γ(x2 ).

(5)

For example, if x1 ≥ 2, x2 ≥ 2, x1 ≤ a, x1 + x2 ≤ a, then relation (5) is true. 1 Now, remark that (5) is not true for all x1 , x2 > 0. Indeed, put x1 = x2 = . 2 µ ¶ √ 1 = π, so if (5) is true, then Γ(1) = 1 ≥ π contradiction. But Then Γ 2 1 (5) is not true, if one of x1 , x2 is ≥ 1. Put e.g. x1 = 1, x2 = to obtain a 2 contradiction. 200

Note that relation (2) can be written also as Γ(x1 + · · · + xn ) ≥

x1 . . . xn Γ(x1 ) . . . Γ(xn ) x1 . . . xn

(6)

if xi > 0, xi ∈ (0, a], x1 + · · · + xn ∈ (0, a].

References [1] M. Petrović, Sur une fonctionnelle, Publ. Math. Univ. Beograde, 1 (1932), 159-156. [2] E. Artin, The Gamma function, Holt, Rinehart and Winston, Inc., New York, 1964.

10

A convexity property of the Gamma function Z Let Γ(x) =

+∞

e−t tx−1 dt (x > 0) be the Euler Gamma function. It is

0

well known that Γ is a convex function for x > 0, and even a stronger result is true: Γ is log-convex function. Let a > 0 be arbitrary positive real number. Then one has ([2]): Theorem 1. Fa (x) = ax Γ(x) (x > 0) is a convex function of x. Proof. It is immediate that Fa0 (x) = ax ln aΓ(x) + ax Γ0 (x),

Fa00 (x) = ax [y 2 Γ(x) + 2yΓ0 (x) + Γ00 (x)],

where y = ln a. The trinom f (y) = y 2 Γ(x)+2yΓ0 (x)+Γ00 (x) has a discriminant ∆ = 4[(Γ0 (x))2 − Γ(x)Γ00 (x)] ≥ 0 by the log-convexity of the gamma function. Since Γ(x) > 0, it follows that f (y) ≥ 0, i.e. Fa is convex indeed. Another proof shows that a more general result holds true. Theorem 2. Let f > 0 be a log-convex function on (0, +∞). Then Fa (x) = ax f (x) is convex for any a > 0. Proof. The log-convexity of f means that log f (αx + βy) ≤ α log f (x) + β log f (y) for all α, β > 0, α + β = 1, x, y ∈ (0, +∞). Thus f (αx + βy) ≤ (f (x))α (f (y))β . Now, Fa (αx + βy) = aαx+βy f (αx + βy) ≤ aαx+βy (f (x))α (f (y))β 201

(1)

by (1). By Young’s inequality (see e.g. [1]) one can write uα v β ≤ αu + βv for all u, v > 0, α, β > 0, α + β = 1. Put u = ax f (x), v = ay f (y). Thus one can write Fa (αx + βy) ≤ αax f (x) + βay f (y) = αFa (x) + βfa (y), which means that Fa is a convex function. For f (x) = Γ(x), Theorem 2 reduces to Theorem 1.

References [1] D.S. Mitronović, Analytic inequalities, Springer Verlag, 1970, [2] J. Sándor, A convexity property of the gamma function (Hungarian), Erdélyi Mat. Lapok (Bra¸sov), 3(2002), no.2, 12; 4(2003), no.1, 11-12.

11

Monotonicity and convexity (concavity) of some functions related to the Gamma function

In what follows we shall study the convexity (concavity) of functions like Γ(x)1/x , Γ(x + 1)1/x , or their quotients; as well as certain monotonicity properties of related functions. The obtained results solve some open problems on the monotonicity of certain sequences related to n! ([7]), but will constitute also generalizations of some known theorems ([3], [4], [8], [9]).

1. Fundamental relations Lemma 1. For x > 0 one has µ 0 ¶ ∞ Γ (x) 0 X 1 K(x) := = . Γ(x) (x + n)2

(1)

n=0

Proof. By the Weierstrass formula one can write: ∞ ³ Y 1 x ´ −x/n = xeγx e . 1+ Γ(x) n n=1

By taking logarithm, after two times differentiation the result follows. The classical theorems on uniform convergence will assure these steps. It is known that one can prove similarly the existence of derivatives of all orders, and to deduce the analycity of ψ(x) := Γ0 (x)/Γ(x) (”digamma function” of Euler). 202

Lemma 2. For x > 1 one has ln(x − 1) < ψ(x) < ln x.

(2)

Proof. By Γ(x + 1) = xΓ(x) and using the theorem of finite increments, one can find θ ∈ (0, 1) such that ln x = Γ0 (x + θ)/Γ(x + θ). By (1) one can write Γ0 (x)/Γ(x) < Γ0 (x + θ)/Γ(x + θ) < Γ0 (x + 1)/Γ(x + 1), and (2) follows. Lemma 3. For x > 1 one has x ln x − x + 1 < ln Γ(x + 1) < (x + 1) ln(x + 1) − x.

(3)

Proof. By using (2), by integration we get (3). Lemma 4. a) If x > 0, one has 1 1 1 < K(x) < + 2 . x x x

(4)

1 1 < K(x) < . x x−1

(5)

b) If x > 1, one has

Proof. It is well-known that f : (0, ∞) → (0, ∞) is a decreasing function, with lim f (x) = 0, then one has x→∞

Z

∞

f (x)dx < a

∞ X

Z

∞

f (n) < f (a) +

f (x)dx, a

n=a

where a is a natural number (see e.g. [9], p.359). By Z ∞ 1 1 dt = 2 (x + t) x 0 we get

∞ X n=0

1 1 < 2+ 2 (x + n) x

Z

∞

0

dt , (x + t)2

so by using (1), we get relation (4). For x > 1, it is easy to see that 1 1 1 + 2 < . x x x−1 203

Lemma 5. One has Z a+1 √ k(a) = ln Γ(x)dx = a ln a − a + ln 2π,

a > 0.

(6)

a

Proof. One has k 0 (a) = ln Γ(a + 1) − ln Γ(a) = ln a; this implies k(a) = Z 1 √ a ln a − a + C. By letting a → 0+, by Raabe’s integral ln Γ(x)dx = ln 2π 0 √ (see [11]), we get C = ln 2π, and so (6) follows. Theorem 1. Let f (x) and g(x) be two functions defined by f (x) = (Γ(x + 1))1/x ,

g(x) =

f (x + 1) . f (x)

Then g(x) is a strictly decreasing function for x > 1. Proof. Let 1 Γ0 (x + 1) ln Γ(x + 1) − h(x) = · . x Γ(x + 1) x2

(7)

This function is connected to many functions which we shall study in what follows. Since f 0 (x) = f (x)h(x), g 0 (x) = (f 0 (x + 1)f (x) − f 0 (x)f (x + 1))/f 2 (x) = g(x)(h(x + 1) − h(x)), it is sufficient to prove the following Lemma 6. For x > 1, h(x) is strictly increasing function. Proof. We consider only x ≥ 6; we shall see that after the improvements of Lemmas 2,3 we may assume x > 1. By Lemma 4, h0 (x)
1, but is strictly x x increasing for x ≥ 6. 204

µ ¶ f (x) 1 f (x) Proof. The derivative of is h(x) − . By taking into account x x x of Lemma 2 and 3, we can prove that µ ¶ 1 2 x h(x) − < x ln(x + 1) − x ln x + 1 < 0. x Put now t(x) = − Then the derivative of Remark. One has

2 2 Γ0 (x + 1) 1 ln Γ(x + 1) + − . x2 x Γ(x + 1) x

f 2 (x) f 2 (x) is t(x) . By Lemma 6, t(x) > 0 for x ≥ 6. x x

1 < Γ(x + 2)1/(x+1) /Γ(x + 1)1/x < 1 +

1 ; x

by replacing x = r (positive integer), one obtains an inequality of [2]. Theorem 3. The function xg(x) is strictly increasing for x > 1. f (x + 1) Proof. (xg(x))0 = [1 + x(h(x + 1) − h(x))]. f (x) By Lemma 2 and 3 one has 1 + x(h(x + 1) − h(x)) >

S , x(x + 1)2

where S = x3 + x2 + 2x + 1 − (2x2 + x) ln(x + 1)/x > 0. Remark. From the above proof it results that for η(x) = h(x + 1) − h(x) one has the inequality η(x) > −

1 , x(x + 1)

which will be used later, too. Theorem 4. For x > 1 we have ¶ µ √ 1 1 ln x − x + ln 2π + , ln Γ(x) > x − 2 12(x + 1) µ ¶ √ 1 1 ln Γ(x) < x − ln x − x + ln 2π + . 2 12(x − 1) 205

(8) (9)

Proof. Let f be a two-times differentiable function. By the mean-value theorem (”trapezium-formula”) Z b f (a) + f (b) (b − a)3 0 f (t)dt = (b − a) − f (ξ), (10) 2 12 a where ξ ∈ (a, b) (see [5]) for f (t) = ln Γ(t), a = x, b = x + 1 we get by (6): a ln a − a + ln

√ 1 Γ(x) 2π = ln − K(ξ), 2 Γ(x + 1)

ξ ∈ (x, x + 1).

By Lemma 4 we get relations (8) and (9). Corollaries. µ ¶ √ 1 2 x+ ln(x + 1) − x − θ for x > 1, θ = 1 − ln 2π Γ(x) > x 2 ln

√ Γ(2x) < x ln x − x + (2x − 1) ln 2 + ln 2π for x > 0. Γ(x)

(11) (12)

Theorem 5. For x > 1 one has ψ(x) > ln x −

1 1 − , 2x 12(x − 1)2

(13)

ψ(x) < ln x −

1 1 − . 2x 12(x + 1)2

(14)

Proof. Apply relation (10) for f (t) = K(t), where the function K is given by (1). Remark. The obtained inequalities improve on the inequalities 3.6.55 (page 228) of [9]. By the use of the above results, now in Lemma 6 for x > 1 one has 1 A(x) < − ln(x + 1) < 0. x

2. Convexity problems Theorem 6. If x > 0, then the function f1 (x) = (Γ(x))1/x is strictly convex. Proof. By applying Lemma 4, and by successive derivation, one obtains · 0 ¸2 Γ (x) ln Γ(x) 00 f1 (x) > f1 (x) − − 1 > 0. Γ(x) x 206

Remark. This method cannot be applied for f (x) since K(x + 1)
0), see relation (5). Theorem 7. For x ≥ 7, f (x) = Γ(x + 1)1/x is strictly concave. Proof. First we need an auxiliary result: 1 1 Lemma 7. K(x) < for x > . (15) 1 2 x− 2 Proof. By the classical Hadamard inequality ([6]) if f is convex, then µ ¶ Z b a+b f (x)dx ≥ (b − a)f . 2 a For the function K this gives ¶ Z a+1 µ 1 1 < K(x)dx = , K a+ 2 a a implying (15). Remark. A generalization of Hadamard’s inequality is obtained in [10]. For the proof of Theorem 7 remark that · ¸ p f (x) Γ0 ln Γ 00 f (x) = − − 1 − 1 − xK(x + 1) · x Γ x ¸ · 0 p ln Γ Γ − − 1 + 1 − xK(x + 1) . · Γ x

The first paranthesis is negative (see the Proof of Theorem 2); for the second one remark that, by using (15), one can write p 1 √ < 1 − xK(x + 1) for x > 1. 2x + 1 Let now H(x) = 1 +

ln Γ(x + 1) Γ0 (x + 1) − . x Γ(x + 1)

By Theorems 4,5 one has H(x)
0).

If x ≥ 7, then one can easily prove that 1 1 1 1 ln(x + 1) + + 2 − 3 − − 2 − 2 ln(x + 1) > . x 6x 2x(x + 1) x 2x 3x But for x ≥ 4 one has 1 1 1 1 < , + + x2 6x3 2x(x + 1) 6x and for x ≥ 6 one has

1 1 , ln(x + 1) < 2x2 2x implying the proof of Lemma 8. 6h(x) K 0 (x + 1) 3K(x + 1) Proof of Theorem 8. h00 (x) = + − . x2 x x2 By the same lines as in the proof of Lemma 4, by using Lemmas 7 and 8 one obtains h00 (x) > (3x + 1)/x3 (x + 1)3 > 0. f (x + 1) is strictly convex for x ≥ 6. f (x) Proof. g 00 (x) = g(x)[(h(x + 1) − h(x))2 + h0 (x + 1) − h0 (x)]. From the convexity of h it follows that g 00 (x) > 0. Lemma 10. If η(x) = h(x + 1) − h(x), then Theorem 9. The function g(x) =

η(x) < −

x3 + x2 − 9x − 3 < 0 for x ≥ 5; 3

(17)

1 for x > 1; x(x + 1)

(18)

η(x) > −

208

η 0 (x)
1. x2 (x + 1)2

(19)

Proof. η(x) < ((x3 + 3x + 1) ln(x + 1) − (x2 + x) ln x − x2 − x)/x2 (x + 1)2 . 1 But ln(x + 1) < + ln x, so x µ ¶ 1 2 η(x) < (2x + 1) ln x − x + 3 + /x2 (x + 1)2 . x If x ≥ 5, then ln x < x/3, and (17) follows. The inequality (18) follows from the Remark of Theorem 3. For (19), by computing η 0 (x), we can use the 1 1 1 inequalities h(x) < , K(x + 1) > and η(x) > − . x x+1 x(x + 1) Theorem 10. The function xg(x) is strictly concave for x ≥ 5. Proof. The second derivative of xg(x) is g(x)(2η(x) + xη 2 (x) + xη 0 (x)) = g(x)G(x). By Lemma 10, G(x)
0, where ψ(y) = denotes Euler’s Γ(y) digamma function. The function h plays a certain role in the study of many 210

properties involving the function f . For example, the following results have been proved in [3]: a) The function f (x)/x is strictly decreasing for x > 1, but (f (x))2 /x is strictly increasing for x ≥ 6; f (x + 1) b) The function g(x) = is strictly decreasing and strictly convex f (x) for x ≥ 6; c) The function xg(x) = xf (x + 1)/f (x) is strictly increasing for x > 1 and strictly concave for x ≥ 5; d) The function f (x) is strictly concave for x > 7; e) The function h(x) is strictly decreasing and strictly convex for x ≥ 6; f) The function f1 (x) = (Γ(x))1/x is strictly convex for x > 0, and logconcave for x ≥ 6. We now prove first that: Theorem 1. lim f (x)/x = 1/e, decreasingly. x→∞

Proof. In view of a) it is sufficient to prove that · ¸ 1 lim ln Γ(x + 1) − ln x = −1. x→∞ x By using Lemma 2 and Lemma 3 from [3], i.e. ln(x − 1) < Γ0 (x)/Γ(x) < ln x for x > 1,

(1)

x ln x − x + 1 < ln Γ(x + 1) < (x + 1) ln(x + 1) − x for x > 1,

(2)

we can write: µ ¶ 1 1 1 1 −1 + < ln Γ(x + 1) − ln x < ln 1 + + ln(x + 1) − 1 for x > 1, x x x x and for x → ∞, the desired result follows. Theorem 2. Let L(x) = f (x)f (x − 1), x ≥ 2. Then 1 lim L(x) = , e

x→∞

decreasingly.

Proof. Since L0 (x) = f 0 (x) − f 0 (x − 1) < 0 for x > 7, by d), clearly b−a L(x) is strictly decreasing for x > 7. Let now L(a, b) = denote the ln b − ln a logarithmic mean of 0 < b < a. This is a mean (see e.g. [7]), so b < L(a, b) < a. 211

(3)

Let b = (Γ(x))1/(x−1) , a = (Γ(x + 1))1/x in (3). Then b < a becomes ln Γ(x) < x ln x, and this is true by (2) for x ≥ 2. Applying (3) we obtain f (x − 1) x f (x) x ln < L(x) < ln . x f (x − 1) x f (x − 1)

(4)

Since x/(x − 1) → 1, by Theorem 1, (4) implies at once Theorem 2. √ a+b Remarks. 1) If we apply ab < L(a, b) < (see [7]), then we get the 2 following improvement of (4): p 1 x f (x) + f (x − 1) x f (x)f (x − 1) ln < L(x) < ln , x f (x − 1) 2x f (x − 1) which can be applied to strengthen certain possible evaluations of the function L(x). 2) A Corollary of Theorem 2 is that p √ 1 n lim [ n+1 (n + 1)! − n!] = . n→∞ e

(5)

1 does not use the monotony (or concavity of f ) e of L, and is based only on (1) and (2). Thus we have obtained a new proof for (5), besides the known proofs in the literature (Traian Lalescu’s sequence). Another proof of L(x) → 1/e as x → ∞ will follow also from: Theorem 3. 1) f (x)h(x) < L(x) < f (x − 1)h(x − 1) for x > 8; 1 2) f (x)h(x) → as x → ∞. e Proof. By Lagrange’s mean value theorem one has The proof of lim L(x) =

L(x) = f (x) − f (x − 1) = f 0 (ξ),

ξ ∈ (x − 1, x).

By d) f 0 is strictly decreasing; and f 0 (ξ) = f (ξ)h(ξ). This proves relation 1) of Theorem 3. For 2) we prove first that: lim xh(x) = 1.

x→∞

By (1), (2), and the definition of h we easily get ln x − ln(x + 1) −

1 1 ln(x + 1) + 1 < xh(x) < ln(x + 1) − ln x + 1 − , x x 212

(6)

which implies (6). Now, writing 1) as f (x) f (x − 1) (xh(x)) < L(x) < (x − 1)h(x − 1), x x−1 and taking into account of (6) and Theorem 1, we obtain 2). 1 Remark. Since xh(x) > [x−1−ln(x+1)] > 0 for x > 3 (ln 4 = 1.38 · · · < x 2 but ln 3 = 1.09 · · · > 1) we have that h(x) > 0 for x ≥ 3. On the other hand, 1 1 since ln(x + 1) − ln x < , we get h(x) < for all x > 1. Actually, these x x inequalities for h(x) are valid for all x > 0. Theorem 4. For all x > 0 we have 0 < h(x)
−1, t 0 and

Z ψ(z) =

∞

[e−t /t − e−tz /(1 − e−t )]dt for Re z > 0.

0

Let x > 0. Then, by the definition of h(x) one can deduce: Z ∞ H(x) = xh(x) = [e−t /(1 − e−t )]s(xt)dt,

(8)

0

where s(xt) = (1 − e−xt )/xt − e−xt . Put xt = u. Then s(u) = (eu −u−1)/ueu > 0 by the well known inequality u e > u + 1. Since e−t /(1 − e−t ) = 1/(et − 1) > 0 for t > 0, the first part of (7) is proved. In our paper [3] we have proved the validity of the right side of (7) for all x > 1 (see the proof of Théorème 2). In fact this was a simple consequence of relations (1) and (2). We now show that this inequality holds also for x ∈ (0, 1]. Let y = ext in the inequality (y − 1)/y < ln y (y > 0). Then s(xt) < (ext − 1)/ext . Let us consider the application l(a) = ax , where a ∈ [1, et ] for t > 0. t 0 By the Lagrange mean-value theorem we have (ext − 1)/(e Z ∞ − 1) = l (ξ) = 1 xξ x−1 ≤ x since x − 1 ≤ 0, ξ > 1. On the other hand, e−xt dt = by a x 0 1 simple integration. Thus by (8) we have H(x) < x · = 1 for all x ∈ (0, 1], x finishing the proof of Theorem 4. 213

References [1] H. Alzer, Refinement of an inequality of G. Benett, Discrete Math., 135(1994), 39-46. [2] J. Sándor, Remark on a function which generalizes the Harmonic series, C.R. Acad. Bulg. Sci. 41(1988), 19-21. [3] J. Sándor, Sur la fonction Gamma, Publ. C.R.M.P. Neuchˆ atel, Série I, 21(1989), 4-7. [4] J. Sándor, Some diophantine equations involving the factorial of a number, Univ. Timi¸soara, Seminar Arghiriade, no.21(1989), 1-4. [5] J. Sándor, L. Tóth, A remark on the Gamma function, Elem. Math. 44(1989), 73-76. atel, [6] J. Sándor, L. Tóth, On some arithmetic products, C.R.M.P. Neuchˆ Série I, 20(1990), 5-8. [7] J. Sándor, On the identric and logarithmic means, Aequationes Math. 40(1990), 261-270.

13

On the Gamma function III

1. In the first two parts we have studied certain basic or monotonicity, convexity properties of functions related to the Euler gamma function ([5], Γ0 (x) [6]). Let ψ be the ”digamma” functions defined by ψ(x) = , x > 0. The Γ(x) 0 ”trigamma” function ψ has been denoted by K in [5]. In what follows we will use this notation here, too, i.e. 0

K(x) = ψ (x) =

∞ X n=0

1 (x + n)2

(1)

(see [5]). Let gt be a function given by gt (x) =

Γ(x + t) Γ(x)

(2)

defined for all x > 0, if t > 0; and for x > −t if t < 0. The aim of this Note is the study of certain new properties of gt and K. A monotonicity property of gt will imply in a unitary way certain inequalities introduced in paper [3]. 214

2. Let t > 0 be given. Theorem 1. The application gt : (0, ∞) → ∞ given by (2) is strictly increasing. Proof. One has gt0 (x) =

Γ0 (x + t)Γ(x) − Γ0 (x)Γ(x + t) Gt (x) = 2 . 2 Γ (x) Γ (x)

Here ·

¸ Γ0 (x + t) Γ0 (x) Gt (x) = Γ(x)Γ(x + t) − = Γ(x)Γ(x + t)[ψ(x + t) − ψ(x)]. Γ(x + t) Γ(x) Since the function ψ is strictly increasing (see (1)), clearly ψ(x + t) > ψ(x) for t > 0. This finishes the proof of Theorem 1. Theorem 2. Let t < 0 be given. Then gt : (−t, ∞) → R given by (2) is strictly decreasing. Proof. Now ψ(x + t) − ψ(x) < 0 since x + t < x, and the above argument finishes the proof. Applications. We now show how these simple results give in a unitary way certain inequalities deduced by other arguments in [3]. 1) Let p, q, m, n > 0 be real numbers such that (p − m)(q − n)( 0. Then Theorem 3.2 of [3] say that Γ(p + n)Γ(q + m)( Γ(p + q)Γ(m + n).

(3)

Let e.g. p > m, q < n. Then gn−q (p + q) > gn−q (m + q) since p + q > m + q. Γ(p + n) Γ(m + n) Therefore > , yielding (3) with ” n, Γ(p + q) Γ(n + q) p > m, by Theorem 2 the sign of inequality is reversed. 2) Let k, p, m > 0 be such that k(p − m − k)( 0. Theorem 3.4 of [3] says that Γ(p)Γ(m)( Γ(p − k)Γ(m + k). Remark that gk (p − k) =

Γ(p) Γ(m + k) > gk (m + k − k) = gk (m) = Γ(p − k) Γ(m)

for k > 0, p > m + k. For k > 0, p < m + k, the inequality is reversed. 3) Let a, b > 0 such that (a − 1)(b − 1)( 0. Then (Theorem 3.6 of [3]) we have Γ(a + b)( Γ(a + 1)Γ(b + 1). 215

Indeed, let a − 1 > 0, b − 1 > 0. Then by Theorem 1, gb−1 (a + 1) > gb−1 (2), Γ(a + 1 + b − 1) Γ(b + 1) so > , giving (5) with ”>” inequality. For a − 1 < 0, Γ(a + 1) Γ(2) b − 1 < 0 by Theorem 2 one has gb−1 (a + 1) < gb−1 (2), giving (5) with ” Γ(x + m)Γ(y + m)

(6)

for x > 0, y > 0, m > 0. This follows at once from gy (x + m) > gy (m). Remarks. The authors in [3] use Hölder’s inequality to prove that ln Γ is convex and that ψ is concave. We note here first that the logarithmic convexity of Γ (Theorem 3.11 of [3]) is well-known, in fact Bohr and Mollerup (see [2]) have shown that Γ is the single logarithmic-convex solution of the functional equation f (x + 1) = xf (x), f (1) = 1. By (ln Γ(x))00 = K(x) > 0 and ψ 00 (x) = ∞ X 1 K 0 (x) = −2 < 0, the convexity and concavity of ln Γ, resp. ψ (x + n)3 n=0 (Theorem 3.13 of [3]), are trivial consequences of (1). Next, the idea of using Hölder’s inequality in the study of properties of gamma function appears in Wendel [7], where it is proved that µ ¶1−a x Γ(x + a) ≤1 (7) ≤ a x+a x Γ(x) for all 0 < a < 1 and x > 0. Indeed, by Hölder’s inequality one has µZ ∞ ¶1/p µZ ∞ ¶1/q Z ∞ p q f (t)g(t)dt ≤ (f (t)) dt (g(t)) dt 0

0

0

µ

¶ 1 1 1 1 + = 1, p > 1 , where p = , q = , f (t) = e−t tx+a−1 , g(t) = p q a 1−a e−t tx−1 . By Γ(x + 1) = xΓ(x) we obtain the right side of (7). The left side of (7) follows by putting a → 1 − a and x → x + a in the right side of (7). An Γ(x + a) interesting consequence of (7) is that lim = 1 (0 < a < 1). x→∞ xa Γ(x) 3. The function K plays an important role in many problems involving the gamma function. In what follows we will deduce certain interesting properties of this function. 1 First we note that, since ψ(x + 1) = ψ(x) + , by differentiation we clearly x have 1 K(x + 1) = K(x) − 2 , x > 0. (8) x 216

µ ¶ 1 1 1 Similarly, from ψ(2x) = ψ(x) + ψ x + + ln 2 (see e.g. [1], relation 2 2 2 6.3.8) we get the identity µ ¶ 1 4K(2x) = K(x) + K x + , x > 0. (9) 2 Finally, we note that from (1) and for 0 < x < 1, K(x) + K(1 − x) =

∞ X

1 π2 = , 2 2 (x + n) sin πx n=−∞

by a known identity, so: K(x) + K(1 − x) =

π2 , sin2 πx

x ∈ (0, 1).

(10)

A similar identity (which in fact, may be deduced from (10), too), is the Gauss identity π , x ∈ (0, 1). (11) Γ(x)Γ(1 − x) = sin πx We now show that the classical Jensen-Hadamard (or Hermite-Hadamard, or Hadamard) inequality for a convex function, as well as its refinements, can give interesting precise results for K. First apply µ (b − a)K

a+b 2

¶

Z
0 one has K(x)K(y) ≥ [K(x) + K(y)]K(x + y).

(16)

Proof. For x = y, (16) reduces to the right side of (15). In the proof of (16) we will use the logarithmic-convexity of the B function of Euler, defined by Z 1

B(x, y) =

tx−1 (1 − t)y−1 dt,

x > 0, y > 0.

0

For a proof of this result via Hölder’s inequality, see [3]. Now, it is anΓ(x)Γ(y) , so ln B(x, y) = ln Γ(x) + other well-known property that B(x, y) = Γ(x + y) ln Γ(y) − ln Γ(x + y) = u(x, y). This function of two arguments being convex on (0, ∞) × (0, ∞), its gradient must be positive semidefinite, i.e. µ 2 ¶µ 2 ¶ ∂2u ∂2u ∂2u ∂ u ∂ u > 0, ∆ = · 2 − ≥ 0. 2 2 ∂x ∂x ∂y ∂x∂y ∂y∂x Here

∂2u = K(x) − K(x + y) > 0, while ∂x2 2

∆ = [ψ 0 (x) − ψ 0 (x + y)][ψ 0 (y) − ψ 0 (x + y)] − ψ 0 (x + y) = K(x)K(y) − K(x)K(x + y) − K(y)K(x + y) (after certain elementary computations). This yields relation (16). 5. The function ψ is strictly concave on (0, ∞). However, we can prove, by using the function K the following result: Theorem 6. The function x 7→ xψ(x), x > 0, is strictly convex. Proof. (xψ(x))0 = ψ(x) + xψ 0 (x), (xψ(x))00 = 2K(x) + xK 0 (x). Now, by using (1) and the derivability of uniform-convergent series, we can write ∞ X 1 , K 0 (x) = −2 (x + n)3 n=0

which gives after a short computation 2K(x) + xK 0 (x) = 2

∞ X n=0

n > 0. (x + n)3

Therefore the above function is strictly convex. 219

Let now 0 < a 0.

(17)

The following result is true: Theorem 7. The application ha,b is a strictly undecreasing, strictly convex function. Γ0 (x + a)Γ(x + b) − Γ(x + a)Γ0 (x + b) Proof. h0a,b (x) = Γ2 (x + b) · ¸ Γ(x + a)Γ(x + b) Γ0 (x + a) Γ0 (x + b) = − Γ2 (x + b) Γ(x + a) Γ(x + b) = ha,b (x)[ψ(x + a) − ψ(x + b)] < 0 since x + a < x + b and ψ is increasing (ψ 0 (x) = K(x) = 0). On the other hand, h00a,b (x) = h0a,b (x)[ψ(x + a) − ψ(x + b)] + ha,b (x)[K(x + a) − K(x + b)] = ha,b {[ψ(x + a) − ψ(x + b)]2 + K(x + a) − K(xb )} > 0 since K(x + a) > K(x + b), K being a decreasing function. Note added in proof. Theorem 6 has been discovered in 1974 by W. Gautschi [Some mean value inequalities for the gamma function, SIAM J. Math. Anal. 5(1974), 282-292]. However, we give here one more proof of this result. From the well known formula ∞

ψ(z) = −γ −

X 1 1 +z z k(z + k) k=1

we get

µ ¶ ∞ X 1 1 ψ = −γ − z + = a(z). z k(kz + 1) k=1

So a00 (z) = 2

∞ X (kz + 1)−3 > 0, k=1

after a little µ ¶calculation. On the other hand, it is known that xψ(x) is con1 vex iff ψ is convex (see e.g. Hardy-Littlewood-Pólya, Inequalities, p.97, x Theorem 119). 220

References [1] M. Abramowitz, I.A. Stegun, Handbook of mathematical functions with formulas, graphs and mathematical tables, National Bureau of Standards, Washington, 1964. [2] E. Artin, Einf¨ uhrung in die Theorie der Gamma function, Teubner, Leipzig, 1931. [3] S. S. Dragomir, R. P. Agarwal, N. S. Barnett, Inequalities for beta and gamma functions via some classical and new integral inequalities, RGMIA Research Report Collection, 2(1999), no.3, 283-333. [4] J. Sándor, Some integral inequalities, Elem. Math. 43(1988), 177-180. [5] J. Sándor, Sur la fonction gamma, Publ. C.R.M.P. Neuchˆ atel, Série I, 21(1989), 4-7. atel, Série [6] J. Sándor, On the gamma function II, Publ. C.R.M.P. Neuchˆ I, 28(1997), 10-12. [7] J. Wendel, Note on the gamma functions, Amer. Math. Monthly, 55(1948), 563-564.

14

On certain inequalities for the ratios of Gamma functions 1. A recent Open Question [1] asks for the proof of the inequalities of type nb−a ≤

Γ(n + b) ≤ (n + 1)b−a , Γ(n + a)

(1)

where n is a positive integer, while 0 ≤ a < b < 1 are given real numbers. We will prove that the right side of (1) is true, even in much better forms, but generally the left side doesn’t hold. Put e.g. n = 1 in the left side of (1). Then, since Γ(1 + x) = xΓ(x), the inequality becomes bΓ(b) ≥ aΓ(a) for 1 1 0 ≤ a Γ = = 0.885 . . . 4 4 4 2 2 2 First we shall prove that: 221

Theorem 1. For all x > 0, 0 < a 0. Indeed, by Hölder’s inequality one has µZ ∞ ¶1/q Z ∞ Z ∞ f (t)g(t)dt ≤ (f (t))p dt1/p (g(t))q 0

0

0

¶ 1 1 1 1 + = 1, p > 1 . Put p = , q = , f (t) = e−t tx+A−1 , g(t) = p q A 1−A e−t tx−1 . Since Z ∞ Γ(x) = e−t tx−1 dt (x > 0), Γ(x + 1) = xΓ(x), µ

0

we obtain the right side of (3). The left side of (3) follows by putting A → 1−A, x → x + A in the right side of (3). Let now x → x + B − A, where b > A > 0 in the right side of (3). Then one gets Γ(x + B) ≤ (x + B − A)A . Γ(x + B − A) By putting here B → b, A → b − a, this yields the right side of relation (2). The left side of (2) follows by the same argument. Remark 1. An interesting consequence of (3) is the limit lim

x→∞

Γ(x + A) = 1, xA Γ(x)

0 < A < 1 fixed.

(4)

Remark 2. Since (x + a)b−a < (x + 1)b−a , from (2), we obtain a refined version of a generalization of right side of inequality (1). Theorem 2. For all x ≥ 1 and 0 < a < b < 1 one has µ ¶ Γ(x + b) 1/(b−a) − 1 −γ/(x+a) (x + a)e < < (x + b)e 2(x+b) (5) Γ(x + a) (where γ is Euler’s constant). 222

Proof. By taking logarithms in (5), one have to prove ln(x + a) −

γ ln Γ(x + b) − ln Γ(x + a) 1 < < ln(x + b) − . x+a b−a 2(x + b)

(6)

Let f (t) = ln Γ(x+t), t ∈ [a, b]. Applying the Lagrange mean value theorem for f , the middle term in (6) becomes f 0 (ξ) =

Γ0 (x + ξ) = Ψ(x + ξ), Γ(x + ξ)

where ξ ∈ (a, b) and Ψ is the Euler ”digamma function” (see e.g. [5]). Now, it is well-known that 1 Ψ(x) < ln x − , x > 1, (7) 2x see e.g. [2], with improvements. On the other hand, M. Vuorinen and his coworkers (see [3]) have proved that Ψ(x) > ln x −

γ , x

x > 1.

(8)

It is well-known also that Ψ is a strictly increasing function on (0, ∞), so Ψ(x + a) < Ψ(x + ξ) < Ψ(x + b), and applying relation (8) for Ψ(x + a), while (7) for Ψ(x + b), one can deduce (6), i.e. Theorem 2 will be proved. Remark. Since (x + b)e−1/2(x+b) < x + b < x + 1, from the right side of (5), one can again obtain a refinement of right side of (1).

References [1] M. Bencze, OQ.1352, Octogon Math. Mag., 12(2004), no.1, 448. [2] J. Sándor, Sur la fonction Gamma, Publ. C.R. Math. Pures Neuchˆ atel, Série I, 21, 1989, 4-7. [3] M. Vuorinen, Particular letter to the author, dated February 14, 1995. [4] J. Wendel, Note on the gamma function, Amer. Math. Monthly, 55 (1948), 563-564. [5] E.T. Whittaker, G.N. Watson, A course of modern analysis, Cambridge Univ. Press, 1969. 223

15

A note on certain inequalities for the Gamma function

1. Introduction The Euler Gamma function Γ is defined for x > 0 by Z ∞ Γ(x) = e−t tx−1 dt. 0

By using a geometrical method, recently C. Alsina and M. S. Tom´ as [1] have proved the following double inequality: Theorem 1. For all x ∈ [0, 1], and all nonnegative integers n one has Γ(1 + x)n 1 ≤ ≤ 1. n! Γ(1 + nx)

(1)

While the interesting method of [1] is geometrical, we will show in what follows that, by certain simple analytical arguments it can be proved that (1) holds true for all real numbers n, and all x ∈ [0, 1]. In fact, this will be a consequence of a monotonicity property. Γ0 (x) (x > 0) be the ”digamma function”. For properties of Let ψ(x) = Γ(x) this function, as well as inequalities, or representation theorems, see e.g. [2], [4], [5], [7]. See also [3] and [6] for a survey of results on the gamma and related functions.

2. Main results Our method is based on the following auxiliary result: Lemma 1. For all x > 0 one has the series representation ψ(x) = −γ + (x − 1)

∞ X k=0

1 . (k + 1)(x + k)

(2)

This is well-known. For proofs, see e.g. [4], [7]. Lemma 2. For all x > 0, and all a ≥ 1 one has ψ(1 + ax) ≥ ψ(1 + x).

224

(3)

Proof. By (2) we can write ψ(1 + ax) ≥ ψ(1 + x) iff −γ + ax

∞ X k=0

∞

X 1 1 ≥ −γ + x . (k + 1)(1 + ax + k) (k + 1)(1 + x + k) k=0

Now, remark that a 1 a−1 − = ≥0 (k + 1)(1 + ax + k) (k + 1)(1 + x + k) (1 + x + k)(1 + ax + k) by a ≥ 1, x > 0, k ≥ 0. Thus inequality (3) is proved. There is equality only for a = 1. We notice that (3) trivially holds true for x = 0 for all a. Theorem 2. For all a ≥ 1, the function f (x) = Γ(1 + x)a /Γ(1 + ax) is a decreasing function of x ≥ 0. Proof. Let g(x) = log f (x) = a log Γ(1 + x) − log Γ(1 + ax). Since g 0 (x) = a[ψ(1 + x) − ψ(1 + ax)], by Lemma 2 we get g 0 (x) ≤ 0, so g is decreasing. This implies the required monotonicity of f . Corollary. For all a ≥ 1 and all x ∈ [0, 1] one has Γ(x + 1)a 1 ≤ ≤ 1. Γ(1 + a) Γ(ax + 1)

(4)

Proof. For x ∈ (0, 1], by Theorem 2, f (1) ≤ f (x) ≤ f (0), which by Γ(1) = Γ(2) = 1 implies (4). For a = n ≥ 1 integer, this yields relation (1).

References [1] C. Alsina, M. S. Tom´ as, A geometrical proof of a new inequality for the gamma function, J. Ineq. Pure Appl. Math., 6(2)(2005), Art.48. [2] E. Artin, The Gamma function, New York, Toronto, London, 1964. [3] D. S. Mitrinović, Analytic inequalities, Springer Verlag, 1970. 225

´ ements de la théorie des fonctions spéciales, [4] A. Nikiforov, V. Ouvarov, El´ Ed. Mir, Moscou, 1976. [5] J. Sándor, Sur la fonction Gamma, Publ. C. Rech. Math. Pures Neuchâtel, Série I, 21(1989), 4-7. [6] J. Sándor, D. S. Mitrinović (in coop. with B. Crstici), Handbook of number theory, Kluwer Acad. Publ., 1996. [7] E. T. Whittaker, G. N. Watson, A course of modern mathematics, Camb. Univ. Press, 1969.

16

Gamma function inequalities and traffic flow

1. Introduction Let Γ be the Euler gamma function, defined for x > 0 by Z ∞ Γ(x) = e−t tx−1 dt. 0

In 1987, J. Lew, J. Frauenthal and N. Keyfitz [2], by studying certain problems of traffic flow, have proved the double inequality ¶ µ ¶ µ ¶ µ 1 1 1 n ≤Γ Γ(n + 1) ≤ 2 Γ n + , (1) 2Γ n + 2 2 2 where n ≥ 1 is a positive integer. See also M. J. Cloud and B. C. Drachman ([1], p.123). We note that, since µ ¶ √ 1 Γ = π, Γ(n) = (n − 1)! 2 and

µ ¶ √ 1 π Γ n+ = n (2n − 1)!!, 2 2

(see e.g. [3]) inequality (1) – having application to traffic flow, reduces to 2≤

2n n! ≤ 2n , (2n − 1)!

(2)

which can be established e.g. by mathematical induction. Our aim in what follows, is to extend (1) for real arguments. In fact, a stronger relation will be obtained. 226

2. Main results First we prove the following Theorem 1. For any x > 0 one has √ Γ(x + 1) ¶≤ x≤ µ 1 Γ x+ 2

r

1 x+ . 2

(3)

For the proof of (3) we need the following auxiliary result due to J. Wendel [4]: Lemma 1. For all 0 < a < 1 and all x > 0 one has ¶1−a µ Γ(x + a) x ≤ a ≤ 1. x+a x Γ(x)

(4)

Proof. Apply the Hölder inequality Z

µZ

∞

∞

¶1/p µZ

∞

q

(g(t))

(f (t)) dt

f (t)g(t)dt ≤

¶1/q ,

0

0

0

where

p

1 1 + = 1 and p > 1. Put p q p=

1 , a

q=

1 , 1−a

f (t) = e−t tx+a−1 ,

g(t) = e−t tx−1 .

By Γ(x + 1) = xΓ(x), we obtain the right side of (4). The left side of follows by putting a → 1 − a, x → x + a in the right side of (4). 1 Apply now Lemma 1 for a = in order to deduce 2 r 1 Γ(x) 1 1 ¶≤ √ ≤ µ x+ . 1 x 2 x Γ x+ 2 √ By multiplying both sides of (5) with x and using xΓ(x) = Γ(x + 1), follows. Lemma 2. For any x ≥ 3/2 one has µ ¶ √ 2 1 x √ < x and 4 > π x + . 2 π 227

(4)

(5)

(3)

(6)

r 2 3 √ ≤ x, the first relation of (6) is trivial. By π < 4, Proof. Since √ < 2 π for the second part of (6) it will be sufficient to prove that 1 4x−1 ≥ x + . 2

(7)

1 3 Let k(x) = 4x−1 − x − , x ≥ . Since k 0 (x) = 4x−1 ln 4 − 1 ≥ 2 ln 4 − 1 > 0, 2 2 µ ¶ 3 3 k is strictly increasing, so k(x) ≥ k = 0, with equality only for x = . 2 2 3 Theorem 2. For all x ≥ one has 2 µ ¶ 1 r Γ x+ √ 2 1 2x 2 ¶ ≤ x+ < √ . √ < x≤ µ (8) 1 2 π π Γ x+ 2 Proof. This follows by Theorem 1 and Lemma 2. Remark. The weaker inequalities in (8) for n ≥ 2 coincide with relation (1).

References [1] M. J. Cloud and B. C. Drachman, Inequalities with applications to engineering, Springer Verlag, 1998. [2] L. Lew, J. Frauenthal and N. Keyfitz, On the average distances in a circular disc, in: Mathematical Modeling: Classroom notes in applied mathematics, Philadelphia, SIAM, 1987. [3] E. T. Whittaker and G. N. Watson, A course of modern analysis, Cambridge Univ. Press, 1969. [4] J. Wendel, Note on the gamma function, Amer. Math. Monthly, 55 (1948), 563-564.

228

Chapter 6

Means and mean value theorems ”... Men pass away, but their deeds abide.” (A.-L. Cauchy)

”... Now... the basic principle of modern mathematics is to achieve a complete fusion of ’geometric’ and ’analytic’ ideas.” (Jean Dieudonné)

229

1

Monotonicity and convexity properties of means

1. Let a, b > 0 be real numbers. The arithmetic and geometric means of a and b are √ a+b A = A(a, b) = and G = G(a, b) = ab. 2 The logarithmic mean L is defined by L = L(a, b) =

b−a , log b − log a

a 6= b;

L(a, a) = a;

while the identric mean is 1 I = I(a, b) = (bb /aa )1/(b−a) , e

(a 6= b),

I(a, a) = a.

For history, results and connection with other means, or applications, see the papers given in the References of the survey paper [1]. The aim of this paper is to study certain properties of a new type of the identric, logarithmic or related means. These properties give monotonicity or convexity results for the above considered means. 2. Let 0 < a < b, and fix the variable b. Then µ ¶ ³ d b b a ´2 L(a, b) = − log − 1 / log . (1) da a a b (We omit the simple computations), so by the known inequality log x < x − 1 (x > 0, x 6= 1), b we have proved that: a Proposition 1. The mean L(a, b) is a strictly increasing function of a, when b is fixed. Consequence 1. The mean L of two variables is a strictly increasing function with respect to each of variables. 3. An analogous simple computation gives · ¸ d log a − log I(a, b) I(a, b) = I(a, b) . (2) da a−b with x :=

Since I is a mean, for 0 < a < b one has a < I(a, b) < b, so from (2) we obtain: Proposition 2. The mean I(a, b) is a strictly increasing function of a, when b is fixed. 230

Consequence 2. The identric mean I of two variables is a strictly increasing function with respect to each of its variables. d2 d2 4. We now calculate 2 L(a, b) and 2 I(a, b). From (1), (2) after certain da da elementary computations one can obtain: · ¸ 2 a−b d2 a+b L(a, b) = (3) − da2 a2 (log a − log b)2 log a − log b 2   2− b (log a − log I(a, b) − 1) 2 d  a I(a, b) = I(a, b)  (4) . da2 (a − b)2 a+b d2 = A(a, b), clearly L(a, b) > 0. By application of 2 da2 b2 b(ab − L2 ) b log I(a, b) we get that the numerator in (4) is 2 − = < 0 by L a aL2 √ G(a, b) = ab < L(ab), which is a known result. Thus, from the above remarks we can state: Proposition 3. The means L and I are strictly concave functions with respect to each variables. 5. As we have seen in paragraph 1 and 3, one can write the equalities: By L(a, b)
− (equivalent to G < L). Thus, via a − b < 0 we get L a I0 L0 < for 0 < a < b. I L I(a, b) Proposition 4. The function a → is a strictly decreasing function L(a, b) for 0 < a < b. Remark. From (8) and (9) we can immediately see that, for a > b the above function is strictly increasing. 6. From the definition of L and I we can deduce that (for 0 < a < b) Here −

L(a, I) =

a−I , log a − log I

which by (5) yields L(a, I) =

a − I I0 · a−b I

(10)

d a−I I(a, b) and I = I(a, b), etc. By 0 < < 1 a corollary of (10) da a−b is the interesting inequality I0 L(a, I) < (11) I which holds true also for a > b. Similarly, from (9) and the analogous identity of (7) we obtain: L0 I (12) log > (a − b) . b L Thus, from the definition of the logarithmic mean, where I 0 =

L(b, I) < Clearly 0
0. 8. We now consider a function closely related to the means L and I. Put r a−b a f (a) = and g(a) = arctg . I(a, b) b It is easy to see that

b , 4AG where A = A(a, b) etc. On the other hand, by (8) we get g 0 (a) =

f 0 (a) =

b . IL

Thus, for the function h, h(a) = f (a) − 4g(a) we have µ ¶ 1 1 0 h (a) = b − L and G < L we get L0 (t) > 0, I 0 (t) > 0 for all t 6= 0. By extending the definition of L and I at t = 0, we have obtained: Proposition 7. The functions t → L(t) and t → I(t) are strictly increasing functions on R. 10. Closely related to the means L and I is the mean S defined by 1

S = S(a, b) = (aa bb ) a+b . By the identity S(a, b) = we get S(t) =

I(a2 , b2 ) , I(a, b)

log S(xt , y t ) = 2I(2t) − I(t), t

so from (15) we can deduce G2 S (t) = 2 2 t L 0

µ ¶ G2 1− 2 A

234

(16)

where G = G(xt , y t ), etc. By extending the definition of S to the whole real line by S(0) = G, we can state: Proposition 8. The function t → S(t) is strictly increasing function on R. I(xt , y t ) 11. Let f (t) = for t 6= 0, f (0) = 1. L(xt , y t ) By logarithming and using relations (15), the following can be proved: f 0 (t) =

f (t) 2 (L − G2 ), tL2

t 6= 0

(17)

where L = L(xt , y t ). Proposition 9. The function t → f (t) defined above is strictly increasing for t > 0 and strictly decreasing for t < 0 (thus t = 0 is the single minimumpoint of this continuous function). 12. Let µ 2 t t ¶ 1t L (x , y ) s(t) = , t 6= 0; s(0) = eG . I(xt , y t ) By log s(t) = 2L(t) − I(t) and from (15) we easily get: µ ¶ 1 s0 (t) I2 G2 = 2 log 2 − 1 + 2 s(t) t L L

(18)

where L = L(xt , y t ) for t 6= 0. By log

I2 G2 − 1 + >0 L2 L2

(see [1]), with the assumption s(0) = eG , we obtain: Proposition 10. The function t → s(t) is strictly increasing on R. L2 (xt , y t ) Corollary. > etG(x,y) for t > 0. I(xt , y t ) d 13. In what follows we will consider derivatives of forms M (a + t, b + t), dt which will be denoted simply by M 0 (where M is a mean). We then will be able to obtain other monotonicity and convexity properties. It is a simple exercise A to see that A0 = 1, G0 = , where G = G(a + t, b + t), etc. Indeed, G p √ 1 √ d 1 1 A G0 = (a + t)(b + t) = · √ b+t+ √ a+t= . dt 2 G a+t 2 b+t Similarly one can show that L0 =

L2 , G2 235

I0 =

I . L

For the mean S the following formula can be deduced: µ ¶ 1 1 0 S =S −k 2 , A A L µ ¶ a−b 2 where k = > 0. Thus 2 (A − I)0 = 1 −

I < 0, L

(G − I)0 =

A I − >0 G L

(since A > I, L > G, so AL > GI); 2(AL − I 2 ) > AL. From AGL < A2 G < I 3 (see [1]) 2 µ ¶ I3 3 3 0 and (G − I ) = 3 AG − we can deduce (G3 − I 3 )0 < 0. These remarks L give: Proposition 11. A − I, G2 − I 2 , G3 − I 3 are strictly decreasing functions, while G − I is strictly increasing on the real line. (Here A = A(a + t, b + t) etc., t ∈ R). For an example of convexity, remark that µ ¶0 µ ¶ I I L2 00 I = = 2 1− 2 , L L G µ 2¶ 2L2 L = 4 (L − A) L00 = 2 G G (G2 − I 2 )0 =

(we omit the details), so we can state: Proposition 12. The functions L and I are strictly concave on the real line. Corollaries. 1. For t ≥ 0 one has: G(a + t, b + t) − I(a + t, b + t) ≥ G(a, b) − I(a, b) G2 (a + t, b + t) − I 2 (a + t, b + t) ≤ G2 (a, b) − I 2 (a, b) A(a + t, b + t) − I(a + t, b + t) ≤ A(a, b) − I(a, b). µ ¶ I(a + t1 , b + t1 ) + I(a + t2 , b + t2 ) t1 + t2 t1 + t2 2. ≤ I a+ ,b + for 2 2 2 all t1 , t2 > 0, a, b > 0. 236

a−b µr ¶ for a 6= b, P (a, a) = a. a 4 arctan −π b This mean has been introduced by Seiffert [2]. It is not difficult to show that µ ¶0 1 P2 −1 0 , . P = = AG P AG µ ¶0 1 −1 Since = , we get: I IL 1 1 Proposition 13. − is a strictly increasing function of t, where P = I P P (a + t, b + t), etc. Indeed, this follows from the know inequality AG < LI. 1 1 1 1 Corollary. − > − for all t > 0, I(a + t, b + t) P (a + t, b + t) I(a, b) P (a, b) a 6= b. Finally, we prove: P Proposition 14. The functions is strictly increasing on R. L Proof. We have µ ¶0 µ ¶ P P 1 P − . = I I AG L 14. Let P = P (a, b) =

Now, it is known that GA < LP ([3]) implying the desired result.

References [1] J. Sándor, V.E.S. Szabó, On certain means, Octogon Math. Mag., 7(1999), no.1, 58-65. [2] H.-J. Seiffert, Problem 887, Nieuw Arch. Wiskunde 12 (Ser.4), 1994, 230231. [3] H.-J. Seiffert, Letter to the author.

2

Logarithmic convexity of the means It and Lt

In paper [3] we have studied the subhomogeneity or logsubhomogeneity (as well as their additive analogue) of certain means, including the identric and logarithmic means. There appeared in a natural way the following functions: f (t) =

log L(xt , y t ) t

and g(t) = 237

log I(xt , y t ) , t

where x, y > 0 while t 6= 0. The t-modification of a mean M is defined by (see e.g. [4]) Mt (x, y) = (M (xt , y t ))1/t . Therefore, f (t) = log Lt (x, y)

and g(t) = log It (x, y),

where L and I are the well-known logarithmic and identric means, defined by L(a, b) =

b−a (b > a > 0), ln b − ln a

L(a, a) = a;

1 I(a, b) = (bb /aa )1/(b−a) (b > a > 0), e In paper [3] we have proved that

I(a, a) = a.

f 0 (t) =

1 g(t) t2

(1)

g 0 (t) =

1 h(t) t2

(2)

and

G2 I where h(t) = 1 − 2 and g(t) = log , where in what follows G = G(xt , y t ), L L etc. Our aim is to prove the following result. Theorem. Lt and It are log-concavepfor t > 0 and log-convex for t < 0. Proof 1. First observe that as G = xt y t , one has 1 G ln G G0 = p . (xt ln x · y t + y t ln y · xt ) = t 2 xt y t Similarly since L(xt , y t ) = L(x, y) we easily get L0 =

xt − y t , t(x − y)

L log I t

(where we have used the fact that log I(a, b) = −2G h (t) = L 0

µ

G0 L − L 0 G L2 238

b ln b − a ln a − 1). Now, b−a

¶ =

2G2 I log , 2 tL G

after using the above established formulae for G0 and L0 . By calculating g 00 (t) =

th0 (t) − 2h(t) , t3

after certain computations we get ¶ µ I 2G2 1 2G2 00 g (t) = 3 log + 2 . t L2 G L −2 I A−L a+b = (where A = A(a, b) = denotes the arithmetic G L 2 mean; for such identities see e.g. [2]), we arrive at µ ¶ 2 G2 A 00 g (t) = 3 −1 . (3) t L3 Since log

Since by a result of Leach-Sholander [1] G2 A < L3 , by (3) we get that < 0 for t > 0 and g 00 (t) > 0 for t < 0. Proof 2. First we calculate I 0 . Let a = xt , b = y t . By

g 00 (t)

log I(a, b) =

b ln b − a ln a −1 b−a

one has

µ 0

I (a, b) = I(a, b)

b ln b − a ln a b−a

¶0 .

Here µ

b ln b − a ln a b−a

¶0

" µ ¶ # ln b − ln a 2 1 b ln b − a ln a − ab , = t b−a b−a

after some elementary (but tedious) calculations, which we omit here. Therefore µ ¶ 1 G2 I 0 = I log I + 1 − 2 . t L Now

µ 00

f (t) = where

L g (t) = I 0

g(t) t2

¶0 =

1 0 (g (t)t − 2g(t)), t3

µ ¶0 µ ¶ µ ¶ I L I 0 L − L0 I 1 G2 = = 1− 2 L I L2 t L 239

(after replacing L0 and I 0 and some computations). Therefore g 0 (t)t − 2g(t) = 1 − In our paper [3] it is proved that log

I2 G2 − log . L2 L2 I2 G2 > 1 − (and this implies the L2 L2

L2 log-subhomogeneity of the mean ). Thus f 00 (t) < 0 for t > 0 and f 00 (t) > 0 I for t < 0. This completes the proof of the theorem.

References [1] E.B. Leach, M.C. Sholander, Extended mean values II, J. Math. Anal. Appl. 92(1983), 207-223. [2] J. Sándor, V.E.S. Szabó, On certain means, Octogon Math. Mag., 7(1999), no.1, 58-65. [3] J. Sándor, On certain subhomogeneous means, Octogon Math. Mag., 8(2000), no.1, 156-160. [4] M.K. Vamanamurthy, M. Vourinen, Inequalities for means, J. Math. Anal. Appl. 183(1994), 155-166.

3

On certain subhomogeneous means

A mean M : R+ × R+ → R+ is called subhomogeneous (of order one) when M (tx, ty) ≤ tM (x, y), for all t ∈ (0, 1] and x, y > 0. Similarly, M is log-subhomogeneous when the property M (xt , y t ) ≤ M t (x, y) holds true for all t ∈ (0, 1] and x, y > 0. We say that M is additively subhomogeneous if the inequality M (x + t, y + t) ≤ t + M (x, y), is valid for all t ≥ 0 and x, y > 0. In this paper we shall study the subhomogenity properties of certain special means, related to the logarithmic, identric and exponential means.

1. Introduction A mean of two positive real numbers is defined as a function M : R+ × R+ → R+ (where R+ = (0, ∞)) with the property: min{x, y} ≤ M (x, y) ≤ max{x, y}, for all x, y ∈ R+ . 240

(1)

Clearly, it follows that M (x, x) = x. The most common example of a mean is the power mean Ap , defined by µ p ¶1 x + yp p Ap (x, y) = , for p 6= 0; 2 √ A0 (x, y) = xy = G(x, y) (the geometric mean). We have: A1 (x, y) = A(x, y) (arithmetic mean); A−1 (x, y) = H(x, y) (harmonic mean); and, as limit cases: A−∞ (x, y) = min{x, y},

A+∞ (x, y) = max{x, y}.

The logarithmic mean is defined by L(x, x) = x,

L(x, y) =

x−y , for x 6= y; log x − log y

and the identric mean by I(x, x) = x;

1 I(x, y) = e

µ

yy xx

¶

1 y−x

, for x 6= y.

For early result, extensions, improvements and references, see [1], [2], [10], [11]. In [22] and [13] the following exponential mean has been studied: E(x, x) = x,

and E(x, y) =

xex − yey − 1 for x 6= y. ex − ey

Most of the used means are homogeneous (of order p), i.e. M (tx, ty) = tp M (x, y), for t > 0. For example A, H and I are homogeneous of order p = 1, while L is homogeneous of order p = 0. There are also log-homogeneous means: M (xt , y t ) = M t (x, y),

t > 0.

For example, the mean G is log-homogeneous. In [10] it is proved that ³x y ´ 1 I(x2 , y 2 ) ≥ I 2 (x, y) and in [13] that E , ≤ E(x, y). These relations 2 2 2 suggest the study of a notion of subhomogeneity. In [13] a mean M is called t-subhomogeneous when M (tx, ty) ≤ tM (x, y), x, y > 0, holds true, and it is 2 shown that for M = E, this holds true for t = . log 8 241

2. Subhomogeneity and log-subhomogeneity The mean Ap is clearly log-subhomogeneous, since it is well known that Ap < Aq for p < q. We now prove the following result: L2 Theorem 1. The means L, I and are log-subhomogeneous. I Proof. First we note that the log-subhomogeneity of L and I follows from a monotonicity property of Leach and Sholander [7] on the general class of means ¶ 1 µ a a−b x − ya b , Sa,b (x, y) = · b a x − yb if a, b ∈ R, x, y > 0 and ab(a − b)(x − y) 6= 0. It is known that S can be extended continuously to the domain {(a, b; x, y) : a, b ∈ R, x, y > 0} and that L(x, y) = S1,0 ; I(x, y) = S1,1 (x, y). Then the log-subhomogeneity of L and I is equivalent to S1,0 (x, y) ≤ St,0 (x, y) and S1,1 (x, y) ≤ St,t (x, y) for t ≥ 1. L2 However, this result cannot be applied to the mean , of L, for x 6= y we I have xt − y t xt − y t L(xt , y t ) = = L(x, y) . t(log x − log y) t(x − y) Since in [10] (relation (13)) it is proved that xt − y t > I t−1 (x, y) for t > 1, t(x − y) by the above identity, we get L(xt , y t ) > L(x, y)I t−1 (x, y) > Lt (x, y), since I>L

(∗)

Thus, we have obtained (in a stronger form) the inequality L(xt , y t ) ≥ Lt (x, y) for t ≥ 1.

(2)

Another proof of (2) is based on the formula µ ¶ 1 I(xt , y t ) d log L(xt , y t ) = 2 log , x 6= y, dt t t L(xt , y t ) which can be deduced after certain elementary computations. Since I > L, log L(xt , y t ) , t > 0, is strictly increasing, implying the function t → t log L(xt , y t ) > log L(x, y) for t > 1, t 242

giving relation (2). A similar simple formula can be deduced for the mean I, too, namely d dt

µ

log I(xT , y t ) t

¶ =

1 t2

µ ¶ G2 (xt , y t ) 1− 2 t t > 0 by L > G. L (x , y )

Thus I(xt , y t ) ≥ I t (x, y) for all t ≥ 1.

(3)

L2 . This is indeed a mean, since by L ≤ I, we I 2 L L2 have ≤ I. On the other hand, it is known that ≥ G. Thus I I We now study the mean

min{a, b} ≤ G(a, b) ≤ L2 (a, b)/I(a, b) ≤ I(a, b) ≤ max{a, b}, giving (1). Let us consider now the function µ h(t) =

L2 (xt , y t ) I(xt , y t )

¶ 1t

,

t ≥ 1,

which has a derivative h(t) h (t) = 2 t 0

µ ¶ I2 G2 log 2 − 1 + 2 , L L

where I = I(xt , y t ), etc. (we omit the simple computations). Now, in [10] (Inequality (21)) the following has been proved: L < Ie

G−L L

,

or with equivalently log

I G >1− . L L

(4)

By (4) we can write log µ by

G −1 L

I2 2G >2− >1− L2 L

µ

G L

¶2 ,

¶2 > 0. Thus h0 (t) > 0, yielding h(t) > h(1) for t > 1. 243

Remark 1. A refinement of (3) can be obtained in the following manner. Let f (t) = I(xt , y t )/L(xt , y t ), Then

t > 0, x 6= y.

· ¸ G2 (xt , y t ) f (t) 1 − 2 t t > 0 by L > G. f (t) = t L (x , y ) 0

So we can write: I(xt , y t )/L(xt , y t ) > I(x, y)/L(x, y), for t > 1, which according to (∗) gives I(xt , y t ) >

L(xt , y t )I(x, y) > I t (x, y), L(x, y)

t > 1.

(5)

Remark 2. By using the function 1

1

s(t) = (L(xt , y t )) t − (I(xt/2 , y t/2 )) t

and applying the same method (using inequality (4)), the following can be obtained: L(xt , y t )/I(xt/2 , y t/2 ) ≤ Lt (x, y)/I t (x1/2 , y 1/2 ), (6) for any t ∈ (0, 1]. Theorem 2. The mean E is subhomogeneous and additively homogeneous. Proof. Let E(tx, ty) g(t) = , t > 1. t Since "µ # ¶ 2 tx − ety 2 (x − y) e g 0 (t) = 2 tx − t2 et(x+y) . (7) t (e − ety )2 x−y (we omit the elementary computations), it is sufficient to prove g 0 (t) > 0 for t > 0. We then can derive E(tx, ty) ≥ tE(x, y) for any t ≥ 1, i.e. the mean E is subhomogeneous. Let et = A > 1. The classical Hadamard inequality (see e.g. [10], [26]) µ ¶ Z x 1 x+y F (t)dt > F x−y y 2 for a convex function F : [x, y] → R, applied to the function F (t) = At , A > 1, gives x+y Ax − Ay > (log A) · A 2 , x−y 244

giving

t(x+y) etx − ety > te 2 ; x−y

thus g 0 (t) > 0, by (7). The additive homogeneity of E is a consequence of the simple equality E(x + t, y + t) = t + E(x, y). Theorem 3. The means L, I, L2 /I, 2I − A, 3I − 2A are additively superhomogeneous, while the mean 2A − I is additively subhomogeneous. Proof. d L2 [L(x + t, y + t) − t] = 2 − 1, dt G d I [I(x + t, y + t) − t] = − 1, dt L · 2 ¸ µ 2 ¶ d L (x + y, y + t) L 2L −t = − 1 − 1, dt I(x + y, y + t) I G2 2I d [2I(x + t, y + t) − A(x + t, y + t) − t] = − 2, dt L d 3I [3I(x + t, y + t) − 2A(x + t, y + t) − t] = − 3, dt L d I [2A(x + t, y + t) − I(x + t, y + t) − t] = 1 − , dt L where in all cases I = I(x+t, y+t), etc. From the know inequalities G < L < I, some of the stated properties are obvious. We note that 2I −A and 3I −2A are A+L means, since L < 2I − A < A, by I > (see [10]) and I < A. Similarly, 2 2A + G I > (see [11]), gives G < 3I − 2A < A. We have to prove only the 3 inequality µ ¶ I 2L2 − 1 >1 L G2 L2 (implying the additive superhomogeneity of ). Since L2 > IG (see [2]), we I G L > . Now have I L µ ¶ G 2L2 2L G −1 = − >1 2 L G G L since 1 +

G L < 2 < 2 by G < L. L G 245

References [1] H. Alzer, Two inequalities for means, C.R. Math. Rep. Acad. Sci. Canada, 9(1987), 11-16. [2] H. Alzer, Ungleichungen f¨ ur Mittelwerte, Arch. Math., Basel, 47(1986), 422-426. [3] J. Arazy et. al., Means and their iterations, Proc. 19th Nordic Congress on Math., Reykjavik, 1984, 191-212, Icelandic Math. Soc., 1985. [4] B.C. Carlson, The logarithmic mean, Amer. Math. Monthly, 79 (1972), 615-618. [5] B.C. Carlson, M. Vuorinen, An inequality of the AGM and the logarithmic mean, SIAM Review, 33(1991), Problem 91-17, 655. [6] C.G. Gauss, Werke III, K. besselschaftend Wissenchaften Göttingen, 1866, 361-371. [7] E.B. Leach, M.C. Scholander, Extended mean values II, Math. Anal. Appl., 92(1983), 207-223. [8] T.P. Lin, The power mean and the logarithmic mean, Amer. Math. Monthly, 81(1974), 879-883. [9] A.O. Pittenger, Inequalities between arithmetic and logarithmic means, Univ. Beograd Publ. Elektr. Fak. Ser. Mat. Fiz. 680(1980), 15-18. [10] J. Sándor, On the identic and logarithmic means, Aequationes Math., 40(1990), 261-270. [11] J. Sándor, A note on some inequalities for means, Arch. Math., 56(1991), 471-473. [12] J. Sándor, Inequalities for means, Proc. 3t Symposium of Math. and Appl. 3-4 nov. 1989 Timi¸soara, 87-90. [13] J. Sándor, Gh. Toader, On some exponential means, Seminar of math. Analysis, Babe¸s-Bolyai Univ. Preprint no.3, 1990,35-40. [14] J. Sándor, On certain inequalities for means, J. Math. Anal. Appl., 189(1995), 602-606; Ibid. II, 199(1996), 629-635. [15] J. Sándor, Two inequalities for means, Int. J. Math. Sci. 18(1995), 621623. 246

[16] J. Sándor, On certain identities for means, Studia Univ. Babe¸s-Bolyai Math., 38(1993), 7-14. [17] J. Sándor, On subhomogeneous means, manuscript. [18] J. Sándor, On the arithmetic-geometric mean of Gauss, Octogon Math. Mag., 7(1999), no.1, 108-115. [19] H.J. Seiffert, Comment to Problem 1365, Math. Mag., 65(1992), 356. [20] K.B. Stolarsky, The power and generalized logarithmic means, Amer. Math. Monthly, 87(1980), 545-548. [21] V.E.S. Szabó, Some weighted inequalities (submitted). [22] Gh. Toader, An exponential mean, Seminar on Math. Analysis, Babe¸sBolyai Univ. Preprint no.5, 1988, 51-54. [23] M.K. Vamanamurthy, M. Vuorinen, Inequalities for means, J. Math. Anal. Appl., 183(1994), 155-166. [24] J. Sándor, V.E.S. Szabó, On an inequality for the sum of infimums of functions, J. Math. Anal. Appl., 204(1996), 646-654. [25] J. Sándor, On means generated by derivatives of functions, Int. J. Math. Ed. Sci. Techn., 28(1997), 129-158. [26] J. Sándor, Gh. Toader, Some general means, Czechoslovak Math. J., 49(124)(1999), no.1, 53-62. [27] J. Sándor, I. Ra¸sa, Inequalities for certain means in two arguments, Nieuw Arch. Wiskunde, 15(1997), 51-55. [28] J. Sándor, On refinements of certain inequalities for means, Arch. Math. Brno, 31(1995), 279-282. [29] J. Sándor, Gh. Toader, I. Ra¸sa, The construction of some new means, General Mathematics, Sibiu, 4(1996), no.1-4, 63-72. [30] Gh. Toader, Some properties of means (submitted). 247

4

On a method of Steiner

1. According to F. Klein [2], the fact that the function x → x1/x (x > 0) attains its maximum at x = e, may the proved also by the use of the logarithmic inequality log t ≤ t − 1, t > 0. (1) x This method is due to J. Steiner, who applied (1) to t := , x > 0. We e immediately get log x 1 ≤ . (2) x e log e 1 , (2) implies the inequality Since = e e x1/x ≤ e1/e ,

(3)

i.e. the above stated property follows. 2. Let now t := 1/t in (1). We get the reverse inequality 1 log t ≥ 1 − , t

t > 0.

(4)

Apply now the Steiner method, by letting t := x(2 − log x) ≤ e, i.e.

µ

e2 x

x in (4). We easily get e

¶x ≤ ee ,

x > 0.

(5)

µ

¶x e2 This shows that the function x → attains its maximum at x = e. x 3. Apply now the Steiner method for means of two arguments. Let A=

a+b , 2

G=

√ ab,

L=

b−a , log b − log a

1 I = (bb /aa )1/(b−a) e

be the arithmetic, geometric, logarithmic, resp. identric means of the numbers 0 < a < b. Also A(a, a) = G(a, a) = L(a, a) = I(a, a) = a. It immediately follows that log I(a, b) =

a log a − b log b a − 1 = + log b − a, a−b L

i.e. log

I a = − 1. b L 248

(6)

Similarly,

I b = − 1. a L By addition of (6) and (7) we get the identity of H.-J. Seiffert: log

log

I A = − 1. G L

(7)

(8)

For identities of this type see also [3]. I Apply now (1) for t = . By (8) we get the inequality G AG ≤ LI, due to H. Alzer [1]. Applying now (4) for t =

(9) I , we obtain G

A G + ≥ 2. L I By using (4) for t =

A , it follows that L µ ¶L µ ¶A I A ≤ . G L

(10)

(11)

The new inequalities (10) and (11) have been deduced therefore by the Steiner method. See also [4], [5] for other applications.

References [1] H. Alzer, Ungleichungen f¨ ur Mittelwerte, Arch. Math., Basel, 47(1986), 422-426. [2] F. Klein, Vorslesungen u ¨ber die Entwicklung der Mathematik im.19. Jahrhundert, I-II, Berlin, Springer Verlag, 1926-27. [3] J. Sándor, On certain identities for means, Studia Univ. Babe¸s-Bolyai, Math., 38(1993), no.4, 7-14. [4] J. Sándor, On an exponential inequality (Romanian), Lucr. Semin. Didactica Mat., 10(1994), 113-118. [5] J. Sándor, On a logarithmic inequality (Romanian), Lucr. Semin. Didactica Mat., 20(2002), 97-100. 249

5

On Karamata’s and Leach-Sholander’s theorems on means

Let L = L(x, y) be the logarithmic mean of the positive real numbers x and y. The following results are frequently used in the theory of means, as well as other parts of Mathematics. L>

x1/3 y + y 1/3 x , x1/3 + y 1/3

and L>

√ 3 G2 A,

x 6= y

(1)

x 6= y

(2)

where G and A denote the geometric, respectively arithmetic means of x, y. J. Karamata [1] used (1) in the solution of a famous problem due to Ramanujan. Inequality (2) has many applications, and is due to E.B. Leach and M.C. Sholander [2]. In what follows we shall prove the up to now apparently unnoticed fact, that Karamata’s inequality (1) is stronger than the Leach-Sholander inequality (2), i.e. s µ ¶ x+y x1/3 y + y 1/3 x √ 3 2 A = 3 xy G . (3) > 2 x1/3 + y 1/3 By putting x = X 3 , y = Y 3 , the inequality (3) becomes µ

XY 3 + Y X 3 X +Y

¶3

µ 3

>X Y

3

X3 + Y 3 2

¶ (4)

i.e. 2(X 2 + Y 2 )3 > (X 3 + Y 3 )(X + Y )3 .

(5)

Now, inequality (5) may be proved via direct calculation, since it is equivalent to (X 3 − Y 3 )(X − Y )3 > 0, which holds true for any X 6= Y (we omit the details). However, the pretty inequality (5) holds in more general terms, if we can remark that by denoting µ Mt = Mt (X, Y ) = then (5) is equivalent to: M2 >

Xt + Y t 2

p M3 M1 . 250

¶1/t ,

(6)

Now, this follows from the known fact (see e.g. [3] for more general results): Mt is a log-concave function of t, for t > 0. This gives log M2 >

log M1 + log M3 , 2

so (6) follows. In fact, by writing that p Mt > Mt−1 Mt+1 ,

t≥2

(7)

one obtains µ

Xt + Y t 2

¶ 2t

µ >

X t−1 + Y t−1 2

1 µ ¶ t−1

X t+1 + Y t+1 2

1 ¶ t+1

(8)

and by substituting x = X t+1 , y = Y t+1 , after certain manipulations, one obtains the following general inequality: i2 h 3t 1 3t t t µ ¶ x 2(t+1) · y 2(t+1) + y 2(t+1) · x 2(t+1) (t+1)(2−t) x+y t(t−1) >2 xy ³ t−1 ´ 1 t−1 2 t−1 t+1 t+1 x +y

(9)

and this finally implies a generalization of (3) for t = 2.

References [1] J. Karamata, Sur quelque problèmes posés par Ramanujan, J. Indian Math. Soc., 24(1960), 343-365. [2] E.B. Leach, M.C. Sholander, Extended mean values II, J. Math. Anal. Appl., 92(1983), 207-223. [3] H. Shniad, On the convexity of mean value functions, Bulletin A.M.S., 54(1948), 770-776.

6

Certain logarithmic means

1. In paper [1] we have introduced the following general means: For a strictly positive, integrable function p : [a, b] → R, let µZ Ap = Ap (a, b) =

b

¶ µZ b ¶ p(x)xdx / p(x)dx ,

a

a

251

(1)

1 1 = 2 = G2p Gp (a, b)

µZ

b

¶ µZ b ¶ (p(x)/x )dx / p(x)dx , 2

a

(2)

a

µZ b ¶ µZ b ¶ 1 1 = = p(x)/xdx / p(x)dx , Lp Lp (a, b) a a µµZ b ¶ µZ b ¶¶ Ip = Ip (a, b) = exp p(x) log xdx / p(x)dx . a

(3) (4)

a

These formulae define the general means Ap , Gp , Lp , Ip of two positive numbers 0 < a < b. In the particular case of p(x) ≡ 1 we reobtain the classical arithmetic, geometric, logarithmic and identric means A, G, L, I. The means defined by (1)-(4) have been applied in the particular case of p(x) = ex to obtain certain exponential means. Their properties have been studied in [3]. The beb − aea − 1 denoted also as E(a, b) has been first introduced in mean Aex = b e − ea [4]. We now take p(x) = log x, x ∈ [a, b] in (1)-(4). We shall obtain certain new logarithmic means. 2. Let us denote for simplicity Al = Al (a, b) = Alog (a, b); etc. A simple integration by parts gives the identity Z b b2 − a2 x log xdx = log I(a2 , b2 ), 4 a since log I(x, y) =

y log y − x log x − 1. y−x

We thus have obtained the identity Al =

a + b log I(a2 , b2 ) · . 4 log I(a, b)

On the other hand, on base of the known identity (see [2]) S(a, b) =

I(a2 , b2 ) , I(a, b)

where S(a, b) = aa/(a+b) bb/(a+b) is a mean introduced and studied also in [1], [2]. Therefore: Proposition 1. µ ¶ A log S A log I(a2 , b2 ) Al = 1+ = · , (5) 2 log I 2 log I(a, b) 252

where A = A(a, b), etc. Similar identities may be found also for Ll and Gl . Let J be a mean defined by (see [1]) µ ¶ 1 1 J = J(a, b) = 1/I , . (6) a b Proposition 2. Ll = L and

log I , log G

(7)

s log I . log J

Gl = G

(8)

Proof. From Z b log x 1 1 dx = (log2 b − log2 a) = (log b − log a) log ab, x 2 2 a and by (3) (and the definitions of classical means) we easily obtain relation (7). From Z b log x dx = (−a log b + b log a − a + b)/(ab) x2 a and by (2) we get µ Gl = G

b log b − a log a − b + a b log a − a log b − a + b

¶1/2 .

Remark now that log I(a, b) = and

µ log J(a, b) = − log I

b log b − a log a −1 b−a 1 1 , a b

¶ =

b log a − a log b + 1, b−a

so relation (8) follows. For the mean Il the things are more complicated, and we shall introduce a new mean, in order to obtain an identity. Theorem 1. Let M = M (a, b) be a mean defined by · ¸ b(log b − 1)2 − a(log a − 1)2 + b − a M = M (a, b) = exp . (9) b(log b − 1) − a(log a − 1) 253

Then log Il = log M + 2 − 2Al .

(10)

Proof. By a partial integration one gets µZ log Il =

b

¶ µZ b ¶ log xdx / log xdx 2

a

a

b2 − a2 log I(a2 , b2 ) 2 . (b − a) log I(a, b)

b log2 b − a log2 a − = Put B=

b log2 b − a log2 a . b log b − a log a − b + a

Then log Il = B − 2Al , by relation (5). Here B is not a mean, but applying the Cauchy mean-value theorem to the functions f (x) = x log2 x, g(x) = x log x − x (x ∈ [a, b]), we get B = log c + 2, where c ∈ (a, b). But B−2=

b(log b − 1)2 − a(log a − 1)2 + b − a , b(log b − 1) − a(log a − 1)

and thus M is a mean, since log M (a, b) = B − 2 = log c, implying M (a, b) = c ∈ (a, b). Since B = 2 + log M (a, b), relation (10) follows. 3. It is known (see [1]) that for all p one has Ap > Ip > Lp > Gp . However, in our case we can obtain more precise informations. These will be based on the above identities, as well as certain known results for the classical means. Theorem 2. One has the following inequalities: Al > A,

(11)

Ll > L,

(12)

log L > L, log G

(13)

Ll < L

log A , log G

(14)

Al > A

log A , log I

(15)

Ll > L

Gl > G, 254

(16)

s log I , log H

GL < G

(17)

Il > I.

(18)

Proof. From the known inequality I(a2 , b2 ) > I 2 (a, b) (see [1]), and by (5) it follows (11). Similarly, by I > G and by (6) we get (12). In fact, since a stronger relation holds true, namely I > L > G, we get the improvement (13) of (12). Relation I < A implies (14). The inequality I(a2 , b2 ) > A2 (see [1]) along with (5) gives (15), which by A > I contains a refinement of (11). The inequality I > J is true, since it is equivalent to L < A. which is well-known. 1 1 we get Therefore, (8) gives (16). From I(x, y) < A(x, y) for x = , y = a b J > H, where H = H(a, b) denotes the harmonic mean of a and b. Finally, by applying the Cauchy-Bunjakovski integral inequality in the form µZ ¶ Z 2

b

1 · log xdx

b

< (b − a)

a

log2 xdx,

a

from (4) we get log Il (a, b) > log I(a, b), so (18) follows. Theorem 3. Al > 1 if and only if I(a, b) > 1. µZ b ¶ µZ b ¶ Z b Proof. Al − 1 = x log xdx − log xdx / log xdx . a

a

(19)

a

Now, remark that x log x − log x = (log x)(x − 1) ≥ 0 for all x > 0, since for x ≥ 1 we have log x ≥ 0, while for 0 < x ≤ 1, log x ≤ 0. Thus the sign of Z b Al − 1 depends on the sign of log xdx. By (4) we get a

1 log I(a, b) = b−a

Z

b

log xdx, a

thus we must have log I(a, b) > 0, i.e. I(a, b) > 1. Remark. The above proof shows also that Al = 1 is impossible. By (19), Al > 1 if e.g. a ≥ 1 (b > a), and 0 < Al < 1, if b ≤ 1. This follows from a < I(a, b) < b. Theorem 4. 2Al −

M 3, M

Al > 1.

(20) (21)

1 Proof. By the logarithmic inequalities log x ≤ x − 1 and log x ≥ 1 − for x M x= , and by using (10) we get relation (20). Since M = Il iff Al = 1, from Il the above Remark, we cannot have equality. Relation (21) is also proved. Remark. The first relation of (20) may be written also as Il + M > Al Il . 2

(22)

4. Finally, we want to state certain Open Problems. It is well-known that in the case of classical means there hold true inequalities like L > G or A > I. What about the inequality Ll > Gl ? By (6) and (8) it is immediately seen, that this is equivalent to the relation µ log I log J >

G log G L

¶2 .

(24)

log Al + 2Al > 2 + log M.

(25)

The inequality Al > Il becomes equivalent to

References [1] J. Sándor, On the identric and logarithmic means, Aequationes Math., 40(1990), 261-270. [2] J. Sándor, I. Ra¸sa, Inequalities for certain means in two arguments, Nieuw Arch Wiskunde, 15(1997), 51-55. [3] J. Sándor, Gh. Toader, Some general means, Czechoslovak Math. J., 49(124)(1999), 53-62. [4] Gh. Toader, An exponential mean, Babe¸s-Bolyai Univ. Preprint 7(1988), 51-54.

7

The arithmetic-geometric mean of Gauss

1. Let x, y be positive real numbers. The arithmetic-geometric mean of Gauss is defined as the common limit of the sequences (xn ), (yn ) defined recurrently by 1 x0 = x, xn+1 = (xn + yn ), n ≥ 0; (1) 2 256

y0 = y, yn+1 =

√ x n yn ,

n ≥ 0.

Let M = M (x, y) = lim xn = lim yn . The mean M was studied firstly n→∞

n→∞

by Gauss [8] and Lagrange [9] (see also [4], [6]), but the real importance of this mean, and its connection with elliptic integrals, is due to Gauss. For historical remarks and a very extensive bibliography on M , see e.g. [4], [2]. See also [16]. The logarithmic and identric means of x and y are defined by L = L(x, y) = (x − y)/(log x − log y) for x 6= y;

L(x, x) = x

(2)

and

1 I = I(x, y) = (xx /y y )1/(x−y) for x 6= y; I(x, x) = x, (3) e respectively. For a survey of results, refinements and extensions, we quote [5], [15], [1], [11], [12]. Very recently, by using a monotone from of l’Hospital’s rule and the representation of M with elliptic integrals, M.K. Vamanamurthy and M. Vuorinen ([16], Theorem 1.3) have obtained among other results the following inequalities for these means: √ (4) M < AL π L<M < L (5) 2 M yn are implied by yn < xn , n ≥ 1, i.e. we can write: 0 < y1 < y2 < · · · < yn < xn < xn−1 < · · · < x1 , 258

n > 1.

(15)

This means that, the sequences (xn ) and (yn ) are convergent. By putting 1 lim xn = u, lim yn = v, from (10) we get u = (u + v), giving u = v(= M ). n→∞ n→∞ 2 We now are turning to the sequences (an ) and (bn ). First we show that an > bn √ for n ≥ 1. Since a1 = (x + y)/2 > b1 = xy, by admitting the property for n, the induction step implies: p 1 an+1 = (an + bn ) > bn+1 = an+1 bn 2 if we can prove that an+1 > bn . This is true, since an+1 = (an + bn )/2 > bn . The monotonicity of (an ) follows immediately: a2 = (a1 + b1 )/2 < a1 and by 1 assuming that an+1 < an , we get: an+2 = (an+1 + bn+1 ) < an+1 , which is 2 equivalent with bn+1 < an+1 . Similarly, the sequence (bn ) is strictly increasing for n ≥ 1. Similarly, the sequence (bn ) is strictly increasing for n ≥ 1. Thus: 0 < b1 < b2 < · · · < bn < an < an−1 < · · · < a1 ,

n > 1.

(16)

The fact that (an ) and (bn ) have a common limit follows exactly by the method used for (xn ) and (yn ). The common limit is the logarithmic mean L of x and y, as proved by Carlson [5]. Corollary 1. For x 6= y one has √ A+G AG < M < 2

(17)

(i.e. (7) and (9)). This follows by the observation that x2 = (x1 + y1 )/2 = (A + G)/2; y2 = √ √ x1 y1 = AG and the fact that xn < x2 for n > 2 and yn > y2 for n > 2. By taking limits in these relations, we get (17) with ”≤” in place of ” y2 . Since µ x3 = one gets:

Ã√ r √ !2 ¶ A+ G A + G√ A+G √ + AG /2 = AG, and y3 = 2 2 2 r

A + G√ AG < M < 2

Ã√ √ !2 A+ G for x 6= y 2

giving an improvement of (17). Theorem 2. The sequence (an + 2bn )n≥1 is strictly increasing. 259

(18)

Proof. Indeed, one has

r 1 1 an+1 + 2bn+1 = (an + bn ) + 2 (an + bn )bn 2 2 1 1 < (an + bn ) + (an + bn ) + bn = an + 2bn 2 2 √ (where we have used 2 αβ < α + β for α 6= β, positive). Corollary 2. The following sequences of inequalities are valid: 1 1 1 A + 2G L ≤ (an + 2bn ) ≤ (an−1 + 2bn−1 ) ≤ · · · ≤ (a1 + 2b1 ) = (19) 3 3 3 3 for all x, y and n ≥ 2. Inequality (19) contains an improvement of (14). For example, for n = 2, after certain easy calculations we get: sµ ¶ A + 2G A+G 2 A+G + G≤ (20) L≤ 6 3 2 3 which seems to be new. Theorem 3. For all n ≥ 1 one has an ≤ xn

and

bn ≤ yn .

(21)

Proof. For n = 1, 2 we have equality; assume that (21) are valid for a fixed n and try to show the validity of (21) for n + 1 in place of n. Since an+1 = (an + bn )/2, by p (21) we can write an+1 ≤ (xn + yn )/2 = xn+1 . One √ the same lines, bn+1 = an+1 bn ≤ xn yn since bn ≤ yn and an+1 ≤ xn by bn ≤ yn ≤ xn . This is true by yn ≤ xn (see (15)). Thus (21) is proved. Remark. For n ≥ 3, one has strict inequalities in (21). Corollary 3. L ≤ M (i.e. (8)), a consequence of (21) and (11), by taking n → ∞. Theorem 4. Let x 6= y, and put un = xn /yn , vn = an /bn (n ≥ 0). Then: a) un > 1, vn > 1 for all n ≥ 1 b) un+1 < un , vn+1 < vn for all n ≥ 1 c) un ≤ vn for n ≥ 1, with equality only for n = 1 d) The sequence (an yn2 /b2n )n≥2 is strictly increasing. A A > 1, v1 = > 1, so the property is true for n = 1. Proof. a) u1 = G G Now let us remark that by (1) one has µr r ¶ 1 xn yn √ xn+1 /yn+1 = (xn + yn )/2 xn yn = + 2 yn xn 260

implying un+1

1 = 2

r ¶ µ √ 1 , un + un

By similar argument, one can derive p vn+1 = (vn + 1)/2,

n = 0, 1, . . .

n = 1, 2, . . .

(22)

(23)

√ Now, we have un+1 > 1, since u + 1 > 2 un if un 6= 1 (we have assumed n r 2 un > 1) and vn+1 > = 1 by (23). Thus, relation a) is proved. 2 b) This is a trivial consequence of Theorem 1. c) ForÃn = 1 one!has equality, but u2 < v2 . First remark that the function r 1 1 f (x) = x+ is strictly increasing for x > 1 (indeed, one has f 0 (x) = 2 x µ ¶ 1 1 √ 1− > 0), so, if we admit that un ≤ vn , than f (un ) ≤ f (vn ) holds x 4 x true by a), i.e. p √ un+1 ≤ f (vn ) = (vn + 1)/2 vn < vn+1 = (vn + 1)/2, since this is equivalent with vn + 1 < 2vn , i.e. vn > 1. This has been proved in part a). By induction it results that c) is valid for all n ≥ 1, with equality only for n = 1. d) Let tn = an yn2 /b2n . Then tn+1 < tn iff an+1 xn yn /an+1 bn < an yn2 /b2n , i.e. xn /yn < an /bn (n > 1), an inequality proved at c). 2 Corollary s 4.µa) xn ≤¶an A, n ≥ 1. A+G b) M 2 ≤ A · L ≤ AL 2 with equality only for x = y. Inequality a) follows from Theorem 4c): x2n ≤ a2n yn2 ≤ an A by an yn2 /b2n ≤ A. Indeed, this is a consequence of Theorem 4d), by tn ≤ t1 = A for n ≥ 1. In sµorder to¶prove b), consider the inequality tn ≤ t3 for n ≥ 3, where t3 = A+G A. By taking limits and using (11), we get b) if we remind that 2 t3 ≤ t2 . Equality can have only for x = y because (tn ) is strictly decreasing for x 6= y. Remark. 1. Inequality b) contains an improvement of (4). 261

2) Remarking that L(x2 , y 2 ) = A(x, y)L(x, y) = AL, and applying Corollary 3 and Corollary 4b), one has the following inequalities: L2 (x, y) ≤ M 2 (x, y) ≤ L(x2 , y 2 )

(24)

M 2 (x, y) ≤ L(x2 , y 2 ) ≤ M (x2 , y 2 ).

(25)

For further properties of similar vein, see [13], where a notion of subhomogenity is introduced. Another algorithm of Borchardt (see [3], [4]) is obtained via the sequences p 1 z0 = x, zn+1 = (zn + tn ); t0 = y, tn+1 = zn+1 tn , n = 0, 1, 2, . . . 2

(26)

Theorem 5. a) For x > y, the sequences (zn )n≥1 and (tn )n≥1 are strictly increasing, and - increasing, respectively. b) For x < y, the sequences (zn )n≥1 and (tn )n≥1 are strictly increasing, and - decreasing, respectively. c) The sequences (zn ) and (tn ) are convergent and have the same limit W = W (x, y). d) For x > y one has zn ≤ an , tn ≤ bn for all n ≥ 1. e) For x < y one has zn ≥ an , tn ≥ bn for all nr ≥ 1. x+y y = t1 . Thus t2 = Proof. For x > y we have z1 = (x + y)/2 > 2 √ z2 t1 > t1 since z2 = (z1 + t1 )/2 > t1 . Assume inductively that tn+1 > tn . √ Then tn+2 = zn+2 tn+1 > tn+1 iff zn+2 > tn+2 , i.e. zn+1 + tn+1 > 2tn+1 , √ which is true by zn+1 > tn+1 = zn+1 tn , i.e. zn+1 > tn if zn > tn is true. This last inequality can be proved by another inductive argument. Here z1 > t1 √ and if zn > tn is valid, then an+1 = (zn + tn )/2 > zn+1 tn iff zn+1 > tn , being equivalent with zn > tn . In a similar way, one has zn+1 < zn , since (zn + tn )/2 < zn is the same as tn < zn . b) For x < y the proof is analogous with that of a). c) In all cases, (zn ) and (tn ) are convergent, and taking limits in (26) one obtains the same limit lim zn = lim tn = W . n→∞

Kn→∞

d) For x > y we have (x + y)/2 < x, i.e. t1 < b1 . Here z1 = a1 . Now, by √ induction, p zn+1 = (zn + tn )/2 ≤ (an + bn )/2 = an+1 and tn+1 = zn+1 tn ≤ bn+1 = an+1 bn since tn ≤ bn is admitted and zn+1 ≤ an+1 has been proved above. e) This is similar with the proof of d). Corollary 5. a) For x ≥ y one has W ≤ M . b) For x ≤ y one has W ≥ M . 262

c) For x ≤ y one has L ≤ W ≤ M . Relations a) and b) are consequences of Theorem 5d), e), by taking n → ∞ in these inequalities; while c) follows via a) and Corollary 3. Remark. The inequality a) in Corollary 5 has been obtained by the author firstly by using the representation of M by Z

1 2 = M (x, y) π

π 2

q dt/ x2 sin2 t + y 2 cos2 t

(27)

0

due to Gauss [8]. By the well-known Jordan inequality sin t ≥

h πi 2 t for t ∈ 0, π 2

(28)

we get from (27) that 2 1/M (x, y) ≤ π

Z

π 2

s dt/

y2

0

+

t2

µ p ¶2 2 2 2 x −y π

and using the integral Z 0

π 2

¯π p 1 ¯2 dx/ a2 + x2 b2 = log(bx + a2 + b2 x2 )¯ , b 0

after certain elementary computations, we arrive at Ã 1/M (x, y) ≤

log

p

x2 − y 2 + x y

!

p / x2 − y 2 .

(29)

The right-hand side of (29) is exactly 1/W (x, y), as is proved in [4]. However, the sequential method of proof of a) is much simpler, as can be seen from the proof of Theorem 5. 4. A remark on inequality (6). Finally, let us remark that the left side of inequality (6) is a simple consequence of (4). Indeed, it is known (see [11] or [12]) that A+L √ I> > AL, 2 so (6) trivially follows from (4). Acknowledgements. The author is indebted to Professors Carlson and Vuorinen for reprints of [4], [5], [7] and a copy of [16]. 263

References [1] H. Alzer, Ungleichungen f¨ ur Mittelwerte, Arch. Math., 47(1986), 422426. [2] J. Arazy, T. Claeson, S. Janson, J. Peetre, Means and their iterations, Proc. 19th Nordic Congress of Math., Reykjavik, 1984, 191-211, Icelandic Math. Soc., 1985. [3] C.W. Borchardt, Gesammelte Werke, Berlin, 1888, 455-462. [4] B. C. Carlson, Algorithms involving arithmetic and geometric means, Amer. Math. Monthly, 78(1971), 496-505. [5] B.C. Carlson, The logarithmic mean, Amer. Math. Monthly, 79 (1972), 615-618. [6] B.C. Carlson, Special functions of Applied Mathematics, Academic Press, New York, 1977. [7] B.C. Carlson, M. Vuorinen, An inequality of the AGM and the logarithmic mean, SIAM Reviews 33(1991), Problem 91-17, 655. [8] C.F. Gauss, Werke, vol.3, Teubner, Leipzig, 1876, 352-355 and Werke, vol.10, part 1, Teubner Leipzig, 1917. [9] J.L. Lagrange, Oeuvres, vol.2, Paris, 1868, 253-312. [10] E.B. Leach, M.C. Sholander, Extended mean values II, J. Math. Anal. Appl., 92(1983), 207-223. [11] J. Sándor, On the identic and logarithmic means, Aequationes Math., 40(1990), 261-270. [12] J. Sándor, A note on some inequalities for means, Arch. Math. Basel, 56(1991), 471-473. [13] J. Sándor, Subhomogenity of means (to appear). [14] J. Sándor, On certain inequalities for means, J. Math. Anal. Appl., 189(1995), 602-606. [15] K.B. Stolarsky, The power mean and generalized logarithmic means, Amer. Math. Monthly, 87(1980), 545-548. 264

[16] M.K. Vamanamurthy, M. Vuorinen, Inequalities for means, J. Math. Anal. Appl., 183(1994), 155-166. Note added in proof. Certain results of these types have been published also in [J. Sándor, On certain inequalities for means II, J. Math. Anal. Appl., 199(1996), 629-635].

8

On two means by Seiffert 1. Let A, G, Q be the classical means of two arguments defined by

x+y √ , G = G(x, y) = xy, 2 r x2 + y 2 , x, y > 0. Q = Q(x, y) = 2 Let L and I denote the logarithmic and identric means. It is well-known that G < L < I < A for x 6= y. In 1993 H.-J. Seiffert [4] introduced the mean A = A(x, y) =

P = P (x, y) =

x−y r (x 6= y), x 4 arctan −π y

P (x, x) = x

and proved that L < P < I for x 6= y. In [5] he obtained other relations, too. The mean P can be written also in the equivalent form P (x, y) =

x−y x−y 2 arcsin x+y

(x 6= y),

(1)

see e.g. [3]. Let x < y. In the paper [1] we have shown that the mean P is the common limit of the two sequences (xn ), (yn ), defined recurrently by x0 = G(x, y),

y0 = A(x, y),

xn+1 =

x n + yn , 2

yn+1 =

√ xn+1 yn .

This algorithm appeared in the works of Pfaff (see [1]). By using simple properties of these sequences, strong inequalities for P can be deduced. For example, in [1] we have proved that xn
0;p and T (u, v) = P (x, y). Proof. From p2(u2 + v 2 ) > |u − v| we get that x > 0, y > 0. Clearly one has x + y = 2(u2 + v 2 ), x − y = u − v. From the definitions (1) and u−v u−v = arcsin p . Let u > v, and put (4) we must prove arctan u+v 2(u2 + v 2 ) u−v α = arctan . By u+v tan α sin α = cos α tan α = √ 1 + tan2 α 266

and

u−v u−v s uµ+ v ¶ = p 2 2(u2 + v 2 ) u−v 1+ u+v

we get u−v u−v arcsin p = α = arctan , 2 2 u+v 2(u + v ) and the proof of the above relation is finished. It is interesting to remark that x+y = A(x, y) = 2 G(x, y) =

r

u2 + v 2 = Q(u, v) 2

u+v √ xy = = A(u, v). 2

Therefore, by using the transformations of Theorem 1, the following transformations of means will be true: G → A,

A → Q,

P → T.

Thus, the inequality G 

Q+A 2

¶2 Q.

(7)

In fact, the following is true: Theorem 2. Let 0 < u < v. Then T = T (u, v) is the common limit of the sequences (un ) and (vn ) defined by u0 = A(u, v), v0 = Q(u, v), un+1 = For all n ≥ 0 one has un < T < vn and 267

u n + vn √ , vn+1 = un+1 vn . 2 p 3

vn2 un < T
0 let us denote A = A(a, b) =

a+b , 2

a2 + b2 W = W (a, b) = , a+b

G = G(a, b) =

√ ab,

H = H(a, b) = 2/

µ

1 1 + a b

¶ .

If f : [a, b] → R is an increasing (decreasing) function, then the following property is immediate: af (b) + bf (a) f (a) + f (b) af (a) + bf (b) Proposition 1. ≤ ≤ (1) a+b 2 a+b All inequalities in (1) are reversed, when f is decreasing. Proof. After simple computations, each parts of (1) become equivalent to (f (b) − f (a))(b − a) ≥ 0 (or (f (b) − f (a))(b − a) ≤ 0). For f (x) = x, relations (1) imply the classical inequality H ≤ A ≤ W. A more interesting example arises, when f (x) = ln x. Then we get (ab ba )1/(a+b) ≤ G ≤ (aa bb )1/(a+b) .

(2)

For the involved means in the extremal sides of (2), see e.g. the References [1]-[3]. 268

If f is convex, the following can be proved: Proposition 2. Let f be convex on [a, b]. Then f (W ) ≤

af (a) + bf (b) ; a+b

(3)

f (H) ≤

af (b) + bf (a) ; a+b

(4)

af (b) + bf (a) + f (W ) ≤ f (a) + f (b). a+b µ Proof. f (W ) = f

≤

a2 + b2 a+b

¶

(5)

µ ¶ a b =f a +b a+b a+b

a b af (a) + bf (b) f (a) + f (b) = , a+b a+b a+b

by the convexity of f (i.e. f (aλ + bµ) ≤ λf (a) + µf (b) for λ, µ > 0, λ + µ = 1). This proved (3). Now, µ ¶ µ ¶ 2ab a b f (H) = f =f b+ a a+b a+b a+b ≤

a b af (b) + bf (a) f (b) + f (a) = , a+b a+b a+b

yielding (4). Relation (5) follows by (3), since af (b) + bf (a) af (a) + bf (b) + = f (a) + f (b). a+b a+b 2. By taking into account of Propositions 1 and 2, one can ask the question of validity of relations of type af (b) + bf (a) af (a) + bf (b) ≤ f (W ) ≤ , a+b a+b or f (H) ≤

af (b) + bf (a) ≤ f (A), etc. a+b

We will prove the following results: 269

Theorem 1. ([4]) Let f : [a, b] → R be a differentiable, convex and increasf 0 (x) ing function. Suppose that the function g(x) = , x ∈ [a, b] is decreasing. x Then one has af (b) + bf (a) f (H) ≤ ≤ f (A). (6) a+b Proof. The left side of (6) is exactly relation (4). Let us write the righthand side of (6) in the form a[f (b) − f (A)] ≤ b[f (A) − f (a)]. By b − A = has

(∗)

b−a = A − a, and by the Lagrange mean value theorem one 2

b−a 0 b−a 0 f (ξ2 ), f (A) − f (a) = f (ξ1 ), 2 2 where ξ1 ∈ (a, A), ξ2 ∈ (A, b). Thus a < ξ1 < ξ2 < b. By f 0 (x) ≥ 0, and f 0 being increasing we get by the monotonicity of g: f (b) − f (A) =

f 0 (a) f 0 (b) ≤ , b a so af 0 (ξ2 ) ≤ af 0 (b) ≤ bf 0 (a) ≤ bf 0 (ξ1 ). This implies relation (∗), i.e. the proof of Theorem 1 is completed. The following theorem has a similar proof: Theorem 2. Let f : [a, b] → R be a differentiable, convex, and increasing √ function. Suppose that the function h(x) = f 0 (x)/ x is decreasing on [a, b]. Then af (b) + bf (a) ≤ f (G). (7) f (H) ≤ a+b For f (x) = x, (7) gives the classical inequality H ≤ G. Theorem 3. Let f : [a, b] → R be a differentiable, convex, and increasing f 0 (x) function. Suppose that the function g(x) = is decreasing on [a, b]. Then x f (A) ≤ f (W ) ≤

f (a) + f (b) . 2

(8)

Proof. The left side of (8) is trivial by A ≤ W and the monotonicity of f . The proof of right side is very similar to the proof of right side of (6). Indeed, W −a=

b(b − a) , a+b

b−W =

270

a(b − a) . a+b

By Lagrange’s mean value theorem one has f (W ) − f (a) =

b(b − a) 0 f (η1 ), a+b

f (b) − f (W ) =

a(b − a) 0 f (η2 ), a+b

where η1 ∈ (a, W ), η2 ∈ (W, b). Now, we can write that af 0 (η1 ) ≤ af 0 (b) ≤ bf 0 (a) ≤ bf 0 (η2 ), so f (W ) − f (a) ≤ f (b) − f (W ), and (8) follows. 3. Finally, we shall prove an integral inequality, which improves on certain known results. Theorem 4. ([4]) If f : [a, b] → R is convex and differentiable, then · ¸ Z b 1 1 af (b) + bf (a) f (a) + f (b) f (x)dx ≤ + f (W ) ≤ . (9) b−a a 2 a+b 2 Proof. Since f is convex, and differentiable, we can write that f (x) − f (y) ≤ (x − y)f 0 (x) for all x, y ∈ [a, b].

(∗∗)

Apply now (∗∗) for y = W and integrate the relation on x ∈ [a, b]: Z b Z b (x − W )f 0 (x)dx. f (x)dx ≤ (b − a)f (W ) + a

a

Here Z

b

Z 0

(x − W )f (x)dx =

b

xf 0 (x)dx − W [f (b) − f (a)]

a

a

Z = bf (b) − af (a) −

b

f (x)dx − W [f (b) − f (a)], a

by partial integration. Thus · ¸ Z b af (b) + bf (a) 2 f (x)dx ≤ (b − a) + (b − a)f (W ), a+b a and the left side of (9) follows. The right hand side inequality of (9) is a consequence of relation (5). Remarks. 1) Relation (9) improves the Hadamard inequality Z b f (a) + f (b) 1 f (x)dx ≤ . b−a a 2 271

2) If the conditions of Theorem 1 are satisfied, the following chain of inequalities holds true: Z b af (b) + bf (a) 1 f (H) ≤ ≤ f (A) ≤ f (x)dx a+b b−a a · ¸ f (a) + f (b) 1 af (b) + bf (a) + f (W ) ≤ . (10) ≤ 2 a+b 2 3) The methods of this paper show that the more general means Wk =

ak + bk ak−1 + bk−1

may be introduced.

References [1] J. Sándor, On the identric and logarithmic means, Aequations Math., 40(1990), 261-270. [2] J. Sándor, On certain integral inequalities, Octogon Math. Mag., 5 (1997), 29-35. [3] J. Sándor, Gh. Toader, On means generated by two positive functions, Octogon Math. Mag., 10(2002), no.1, 70-73. [4] J. Sándor, On certain new inequalities of convex functions (Hungarian), Erdélyi Mat. Lapok, 5(2003), no.1, 29-34.

10

On an inequality of Sierpinski on the arithmetic, geometric and harmonic means

1. Let xi , i = 1, n, be strictly positive numbers and denote their usual arithmetic, geometric and harmonic mean by n X

An (x) =

Ã

xi

i=1

n

,

Gn (x) =

n Y i=1

!1

n

xi

,

Hn (x) =

n n X i=1

where x = (x1 , . . . , xn ). 272

1 xi

,

In 1909 W. Sierpinski ([5]) discovered the following double-inequality: (Hn (x))n−1 An (x) ≤ (Gn (x))n ≤ (An (x))n−1 Hn (x).

(1)

The aim of this note is to obtain a very short proof for (1) (in fact, a generalization), by using Maclaurin’s theorem for elementary symmetric functions. For another idea of proof for (1) (due to the present author), which leads also to a refinement of an inequality of Ky Fan, see [1]. For application of (1) see [4]. Now we state Maclaurin’s theorem as the following: Lemma. Let cr be the r-th elementary symmetric function of the x (i.e. the sum of the products, r at a time, of different xi ) and pr the average of these products, i.e. cr pr = µ ¶ . n r Then

1

1

1

1

n−k p1 ≥ p22 ≥ p33 ≥ · · · ≥ pn−k ≥ · · · ≥ pnn .

(2)

See [2], [3] for a proof and history of this result. 2. Our result is contained in the following Theorem. Let k = 1, 2, . . . and define the k-harmonic mean of x by µ ¶ n k Hn,k (x) = X . 1 x1 . . . xk (nk) Then one has the inequality (Gn (x))n ≤ (An (x))n−k Hn,k (x).

Remark. Letting x →

1 , we get x

(Gn (x))n ≥ (Hn,k (x))n−k An,k (x), where

X An,k (x) = pk = 273

x1 . . . xk Cnk

.

(3)

1 n−k Proof. Apply p1 ≥ pn−k from (2), where

pn−k

X x1 . . . xn−k µ ¶ = = n n−k

Ã

n Y i=1

! xi

X   

 1 x1 .¶. . xk  , µ  n k

and we easily get (3). For µ k = 1 one ¶reobtains the right side of inequality (1). By replacing x by 1 1 1 = ,..., , and remarking that x x1 xn µ ¶ µ ¶ µ ¶ 1 1 1 1 1 1 Gn = , An = , Hn = , x Gn (x) x Hn (x) x An (x) we immediately get the left side of (1) from the right side of this relation. This finishes the proof of (1).

References [1] H. Alzer, Versch¨ arfung einer Ungleichung vom Ky Fan, Aequationes Math., 36(1988), 246-250. [2] G.H. Hardy, J.E. Littlewood, G. Pólya, Inequalities, 2nd ed., Cambridge Univ. Press, Cambridge, 1959, Theorem 52. [3] C. Maclaurin, A second letter to Martin Folkes, Esq., concerning the roots of equations, with the demonstration of other rules in algebra, Phil. Transactions, 36(1729), 59-96. [4] J. Sándor, On an inequality of Ky Fan, Babe¸s-Bolyai Univ., Seminar on Math. Analysis, no.7, 1990, 29-34. [5] W. Sierpinski, Sur une inégalité pour la moyenne arithmétique, géometrique et harmonique, Warsch. Sitzunsber. 2(1909), 354-357.

11

An application of Rolle’s theorem

1. Rolle’s theorem from the differential calculus asserts that for a continuous function f : [a, b] → R which is differentiable on (a, b) and f (a) = f (b), there exists at least a θ ∈ (a, b) such that f 0 (θ) = 0. As it is well-known, this 274

theorem implies among others the classical mean-value theorems by Lagrange, Cauchy or Taylor. The aim of this note is to consider a less known application of this theorem, with interesting consequences. Two applications relating to the logarithmic mean of numbers, and the Euler gamma function will be given. 2. Let f : [0, 1] → R be a 3-times differentiable function and define Fk : [0, 1] → R by 1 Fk (x) = f (x) − f (0) − x[f 0 (x) + f 0 (0)] 2 ½ ¾ 1 k 0 0 −x f (1) − f (0) − [f (1) + f (0)] (1) 2 where k ∈ R is a fixed number. Clearly, one has Fk (0) = Fk (1) = 0, thus by Rolle’s theorem there exists θ1 ∈ (0, 1) with Fk0 (θ1 ) = 0. Since 1 1 Fk0 (x) = f 0 (x) − [f 0 (x) + f 0 (0)] − xf 00 (x) 2 2 ½ ¾ 1 0 k−1 0 −kx f (1) − f (0) − [f (0) + f (1)] 2

(2)

we have Fk0 (0) = 0. Applying once again the Rolle theorem, one finds a θ2 ∈ (0, θ1 ) with Fk00 (θ2 ) = 0. By ½ ¾ 1 1 Fk00 (x) = xf 00 (x) − k(k − 1)xk−2 f (1) − f (0) − [f 0 (0) + f 0 (1)] (3) 2 2 we get the following result: If f : [0, 1] → R satisfies the above conditions, and k ∈ R is given, then there exists θ = θ2 ∈ (0, 1) such that 1 f 00 (θ) f (1) = f (0) + [f 0 (0) + f 0 (1)] − . 2 2k(k − 1)θk−3

(4)

Let now f (x) = g[(b − a)x + a], where x ∈ [0, 1] and a < b, being 3-times differentiable etc. function. Applying (4) for this function, one can derive: If g : [a, b] → R is as above and k ∈ R is given, then there exists θ ∈ (0, 1) such that 1 g 000 (ξ)(b − a)3 g(b) = g(a) + [g 0 (a) + g 0 (b)](b − a) − , 2 2k(k − 1)θk−3

(5)

where ξ = (b − a)θ + a. For k = 3 one gets a beautiful result: 1 g 000 (ξ)(b − a)3 g(b) = g(a) + [g 0 (a) + g 0 (b)](b − a) − , 2 12 275

ξ ∈ (a, b).

(6)

If we select

Z

x

g(x) =

F (t)dt, a

where F : [a, b] → R has a continuous second order derivative, then from (6) we can obtain: Z b F 00 (ξ)(b − a)3 b−a F (t)dt = [F (a) + F (b)] − , ξ ∈ (a, b), (7) 2 12 a called also as the trapezium formula (see e.g. [3]). 3. As a first application, let F (t) = log t in (7) and 0 < a < b. After some elementary transformation, we can obtain: (b − a)2 (b − a)2 A(a, b) < + 1 < + 1, 12b2 L(a, b) 12a2 where A(a, b) =

a+b 2

and L(a, b) =

(8)

a−b log a − log b

denote the arithmetic and logarithmic mean, respectively, of a and b. We note that the left side of (8) refines the known relation L(a, b) < A(a, b) (see [2], [7]). For another application, let b = x+1, a = x, where x > 1 and F (t) = Ψ(t), where Ψ is the Euler ”digamma function”, i.e. Ψ(t) =

Γ0 (t) , Γ(t)

with Γ the Euler gamma function. Since 00

Ψ (t) = −2

∞ X n=0

1 (n + x)3

(see [1]), and from the inequalities Z

∞

h(t)dt < 0

∞ X

Z h(n) < h(0) +

∞

h(t)dt, 0

n=0

which are valid for all functions h : (0, ∞) → R+ which are strictly decreasing with lim h(x) = 0 (see [4]), one can derive that x→∞

log x −

1 1 1 1 − < Ψ(x) < log x − − , x > 1. 2 2x 12(x − 1) 2x 12(x + 1)2 276

(9)

which improve certain known relations for Ψ (see also [1], [4]). This application appears also in [5]. 4. Of course, there are various applications of (5), (6), (7). By the same method, the following general result of form (4) can be proved: Let f : [a, b] → R be a 3-times differentiable function, and let α, β, γ, δ, k, γ 6= 0, be given real numbers. Assume that the following conditions hold true: (i) (α − 2β)f 0 (0) = 0; (ii) [f (1) − f (0)](α − γ) + [f 0 (1) + f 0 (0)] + [f 0 (1) + f 0 (0)](γδ − β) = 0. Then there exists θ ∈ (0, 1) such that f (1) = f (0) + δ[f 0 (0) + f 0 (1)] +

α − 2β f 00 (θ) β f 000 (θ) · k−2 − · k−3 . k(k − 1)δ θ k(k − 1)γ θ

(10)

Proof. Consider the function F : [0, 1] → R defined by F (x) = α[f (x) − f (0)] − βx[f 0 (x) + f 0 (0)] −γxk {[f (1) − f (0)] − δ[f 0 (1) + f 0 (0)]}

(11)

and apply the same argument as in the proof of (4) (we omit the details). For various selections of α, β, γ, δ, k and f we can obtain different meanvalue theorems with many possible applications.

References [1] E. Artin, The Gamma function, Holt, Einchart, Winston, New York, 1964. [2] B.C. Carlson, The logarithmic mean, Amer. Math. Monthly, 79 (1972), 615-618. [3] B.P. Demidovich, I.A. Maron, Computational Mathematics, Mir Publishers, Moscow, 1981. [4] D.S. Mitrinović, Analytic Inequalities, Springer Verlag, 1970. [5] J. Sándor, Remark on a function which generalizes the harmonic series, C.R. Acad. Buld. Sci., 41(1988), 19-21. [6] J. Sándor, Sur la function Gamma, Publ. Centre Rech. Math. Pures, Neuchâtel, Série I, 21, 1989, 4-7. [7] J. Sándor, Some integral inequalities, Elem. Math., 43(1988), 177-180. 277

12

An application of Lagrange’s mean value theorem for the computation of a limit

We must determine f : R∗+ → R∗+ so that f (x) → ∞ and x[f (x + 1) − f (x)] → a ∈ R (x → ∞). For a = 0 this is OQ.1074 ([1]). We note here that when f is differentiable, the following is true: If lim xf 0 (x) = a, (1) x→∞

then lim x[f (x + 1) − f (x)] = a.

x→∞

(2)

Indeed, by the Lagrange mean-value theorem one can write f (x + 1) − f (x) = (x + 1 − x)f 0 (ξ) for certain x < ξ < x + 1. x x x < 1 < + 1 it follows →1 ξ ξ ξ x as x → ∞. By x[f (x + 1) − f (x)] = f 0 (ξ) = [ξf 0 (ξ)] and the assumption ξ ξf 0 (ξ) → a as x → ∞ (i.e. ξ → ∞), we get the result (2). Now, as x → ∞, clearly ξ → 0, and by

References [1] M. Bencze, OQ.1074, Octogon Math. Mag., 10(2002), no.2, 1064.

13

An inequality of Alzer, as an application of Cauchy’s mean value theorem The inequality of Alzer [1] is the following: n ≤ n+1

Ã

n

n+1

i=1

i=1

1X r 1 X r i / i n n+1

!1/r (1)

where r ≥ 0 is a real number, while n is a positive integer. In the first part [3] we have obtained an easy proof based on mathematical induction and Cauchy’s mean value theorem. Recently, Chen and Qi [2] discovered the interesting fact that (1) is true also for r < 0. They use a complicated function and Jensen’s inequality (and, of course, mathematical induction). We now simply prove that here one can apply Cauchy’s mean value theorem, again. 278

Indeed, the mathematical induction process (see [3], [2]) leads to the inequality (k + 1)s [(k + 1)1−s − k 1−s ] > (k + 2)s [(k + 2)1−s − (k + 1)1−s ].

(2)

Now, let f, g : [k, k + 1] → R be given by f (x) = (x + 1)1−s ,

g(x) = x1−s

where 0 < s < 1. Remark that by the Cauchy mean value theorem one can write: f 0 (ξ) f (k + 1) − f (k) = 0 with ξ ∈ (k, k + 1). g(k + 1) − g(k) g (ξ) Since

f 0 (ξ) = g 0 (ξ)

µ

ξ ξ+1

¶s

µ
0, ∀ x ∈ (a, b).

(13)

Applying (1) and remarking that F (b) − F (a) = f (b) − f (a),

G(b) − G(a) = g(b) − g(a)

one gets the following result: Let f, g, u, v be Rolle functions on [a, b] with u(a) = a, u(b) = v(a) = c, c ∈ (a, b), v(b) = b and u0 (x), v 0 (x) > 0 and (g ◦ u)0 (x) + (g ◦ v)0 (x) 6= 0, ∀ x ∈ (a, b). Then there exists ξ ∈ (a, b) such that f (b) − f (a) u0 (ξ)f 0 (ξ1 ) + v 0 (ξ)f 0 (ξ2 ) = 0 g(b) − g(a) u (ξ)g 0 (ξ1 ) + v 0 (ξ)g 0 (ξ2 )

(14)

where ξ1 = u(ξ) ∈ (a, c), ξ2 = v(ξ) ∈ (c, b). For the particular case of u(x) =

x+a , 2

v(x) =

x+b 2

√ ax,

v(x) =

√ xb

respectively u(x) =

for 0 ≤ a < b one gets the corollaries: 1. Let f, g be Rolle functions on [a, b] with g 0 (x) + g 0 (y) 6= 0, ∀ x, y ∈ (a, b), µ Then there exist ξ1 ∈ such that

x−y =

b−a . 2

¶ µ ¶ a+b b−a a+b , b , ξ2 ∈ a, and ξ1 − ξ2 = , 2 2 2

f (b) − f (a) f 0 (ξ1 ) + f 0 (ξ2 ) = 0 . g(b) − g(a) g (ξ1 ) + g 0 (ξ2 )

(15)

2. Let f, g be Rolle functions on [a, b], 0 ≤ a < b with √ √ 0 ag (x) + bg 0 (y) 6= 0, ∀ x, y ∈ (a, b) r r √ √ x a ξ1 a with = . Then there exists ξ1 ∈ (a, ab), ξ2 ∈ ( ab, b), = , such y b ξ2 b that √ √ 0 f (b) − f (a) af (ξ1 ) + bf 0 (ξ2 ) √ = √ . (16) g(b) − g(a) ag 0 (ξ1 ) + bg 0 (ξ2 ) 286

For g(x) = x, x ∈ [a, b], and comparing (15) with Lagrange’s mean value theorem applied to the function f , one gets theµ ¶ ¶ µ a+b a+b Corollary. There exist θ ∈ (a, b), ξ1 ∈ , b , ξ2 ∈ a, , 2 2 b−a ξ1 − ξ2 = such that 2 f 0 (θ) =

f 0 (ξ1 ) + f 0 (ξ2 ) . 2

(17)

We note here that f 0 doesn’t need to be continuous (but clearly having Darboux’s property), in the case of continuous functions, relation (17) has been proved in [1]. Let now H, T : [a, b] → R be defined by H = f ◦ u − f ◦ v,

T = g ◦ u − g ◦ v,

where f, g, u, v are Rolle functions, satisfying the conditions (13) and (g ◦ u)0 (x) − (g ◦ v)0 (x) 6= 0 on (a, b). Then applying (1) to the functions H, T one easy obtain the following result: There exists ξ ∈ (a, b) such that u0 (ξ)f 0 (ξ1 ) − v 0 (ξ)f 0 (ξ2 ) f (b) − 2f (c) + f (a) = 0 g(b) − 2g(c) + g(a) u (ξ)g 0 (ξ1 ) − v 0 (ξ)g 0 (ξ2 )

(18)

where ξ1 = u(ξ) ∈ (a, c), ξ2 = v(ξ) ∈ (c, b). For the particular case of u(x) = x+a x+b , v(x) = and g 0 (x) injective (and f, g have a second order derivative 2 2 on (a, b)) one gets µ ¶ µ ¶ µ ¶ ξ+b ξ+a a+b 0 0 µ ¶ f (b) − 2f + f (a) f −f f 00 (θ) 2 2 2 µ ¶ µ ¶ µ ¶ = = 00 (19) a+b ξ+b ξ+a g (θ) g(b) − 2g + g(a) g0 − g0 2 2 2 where ξ ∈ (a, b), θ ∈ (a, b). For the particular case of g(x) = x2 , x ∈ [a, b], relation (19) yields that µ ¶ a+b (b − a)2 00 f (b) − 2f + f (a) = f (θ), θ ∈ (a, b) (20) 2 4 287

when f has a second order derivative on (a, b). To demonstrate the power of this simple relation, put f (x) = ln x, x ∈ [a, b], where 0 < a < b. Then simple calculations give the double inequality A (b − a)2 1 (b − a)2 1 · 2 < ln < · 2 8 b G 8 a

(21)

where

√ a+b , G = G(a, b) = ab 2 denote the arithmetic and geometric means of a and b, respectively. We note that for a = n, b = n + 1, n > 0 from (21) we obtain the logarithmic inequality A = A(a, b) =

1 2n + 1 1 < ln p < 2, 2 8(n + 1) 8n 2 n(n + 1)

n > 0.

(22)

5. Finally, we will consider certain mean value results and inequalities for Riemann integrals. For simplicity we apply (1) for g(x) = x, x ∈ [a, b]. Let Z u(x) Z v(x) F (x) = f (t)dt + f (t)dt a

b

where f : [a, b] → R is a continuous function and u, v : [a, b] → R are Rolle functions satisfying the conditions u(a) = a, u(b) = v(a) = c, c ∈ (a, b), v(b) = b. Since

Z F (b) − F (a) =

b

f (t)dt, a

one obtains that there exists ξ ∈ (a, b) such that Z b f (t)dt = (b − a)[u0 (ξ)f (u(ξ)) + v 0 (ξ)f (v(ξ))].

(23)

a

For u(x) =

a+x b+x , v(x) = one gets 2 2 · µ ¶ µ ¶¸ Z b b−a a+ξ b+ξ f (t)dt = f +f . 2 2 2 a

(24)

For particular functions f one can obtain interesting results from (24). A result of different type can be proved by considering the application Z x+b Z x+a 2 2 F (x) = f (t)dt − f (t)dt (25) a

a

288

where, as above, f is continuous on [a, b]. Since Z F (b) − F (a) =

Z

b a+b 2

f (t)dt −

a+b 2

f (t)dt,

a

we have · µ ¶ µ ¶¸ ξ+b ξ+a f (t)dt − f (t)dt = (b − a) f −f , ξ ∈ (a, b). a+b 2 2 a 2 (26) a+ξ Let us suppose that f is a convex function on [a, b]. Let ξ1 = , 2 µ µ ¶¶ a+b a+b b+ξ . Let A(a, f (a)), X(ξ1 , f (ξ1 )), M ,f , Y (ξ2 , f (ξ2 )), ξ2 = 2 2 2 B(b, f (b)). Since f is a convex function, it is well known that (see e.g. [5]) Z

b

Z

a+b 2

slopeAX ≤ slopeAM ≤ slopeXM ≤ slopeXY ≤ slopeY M ≤ slopeM B ≤ slopeY B thus in particular slopeAX ≤ slopeY M

(27)

and slopeXM ≤ slopeY B i.e.

µ

a+b f (ξ2 ) − f f (ξ1 − f (a) 2 ≤ a+b ξ1 − a ξ2 − 2

and

µ f

¶

¶ a+b − f (ξ1 ) f (b) − f (ξ2 ) 2 . ≤ a+b b − ξ2 − ξ1 2

a+b a+b and − ξ1 = b − ξ2 we get that 2 2 µ ¶ µ ¶ a+b a+b f − f (a) ≤ f (ξ2 ) − f (ξ1 ) ≤ f (b) − f . 2 2

Since ξ1 − a = ξ2 −

289

(28)

By comparing (26) and (28), we get the following unusual double inequality for a convex (continuous) function f : [a, b] → R: ¶ ¸ Z b · µ Z a+b 2 a+b − f (a) ≤ f (t)dt (b − a) f f (t)dt − a+b 2 a 2

· µ ¶¸ a+b ≤ (b − a) f (b) − f . 2

(29)

Clearly, when f is strictly convex, one has strict inequalities in (29). Remark that (29) is an improvement, involving integrals, for the classical inequality for convex functions µ ¶ a+b 2f ≤ f (a) + f (b). (30) 2

References [1] J. Sándor, Problem T36 (Hungarian), Mat. Lapok, Cluj, XCVII, 1992, no.6, 239. [2] J. Sándor, A note on the functions σk (n) and ϕk (n), Studia Univ. Babe¸sBolyai, Mathematica, XXXV, 2, 1990, 3-6. [3] J. Sándor, On an inequality of Alzer, J. Math. Anal. Appl., 192(1995), 1034-1035. [4] J. Sándor, On a mean value theorem, Octogon Math. Mag., 3(1995), no.2, 47-49. [5] A.W. Roberts, D.E. Varberg, Convex functions, Academic Press, 1973.

16

The second mean value theorem of integral calculus

1. The first mean value theorem for Riemann integrals can be stated as follows: Theorem 1. If f, g are integrable on [a, b], g having a constant sign, then there exist η ∈ [m, M ] (where m = inf f (x), M = sup f (x)) such that Z

Z

b

f (x)g(x)dx = η a

g(x)dx. a

290

b

(1)

If the function f is continuous, then there exists ξ ∈ [a, b] such that η = f (ξ). The second mean value theorem for integrals is less known, and as we shall see, it appears in various places under various conditions and different names. Theorem 2. If f, g are integrable on [a, b], and f is monotone, then there exists ξ ∈ [a, b] such that Z

Z

b

a

Z

ξ

f (x)g(x)dx = f (a)

b

g(x)dx + f (b) a

g(x)dx.

(2)

ξ

It will be sufficient to consider only the following theorem, known also as ”Bonnet’s formula” (P.O. Bonnet (1819-1892) French mathematician). Theorem 3. Let f, g be defined on [a, b], with integrable g, and f a positive, monotone decreasing function. Then there exists ξ ∈ [a, b] such that Z

Z

b

ξ

f (x)g(x)dx = f (a) a

g(x)dx.

(3)

a

Indeed, if (3) is true, let us apply it for the function F defined by F (x) = f (x) − f (b), f being decreasing function. One has F (x) ≥ 0, F decreasing, so by (3) we get Z b Z ξ F (x)g(x)dx = F (a) g(x)dx, a

i.e.

Z

a

Z

b

a

Z

b

f (x)g(x)dx −

f (b)g(x)dx = (f (a) − f (b)) a

Thus Z b

ξ

g(x)dx. a

Z

·Z

ξ

f (x)g(x)dx = f (a)

g(x)dx + f (b)

a

g(x)dx −

a

Z = f (a)

Z

b

a

Z

ξ

g(x)f x + f (b) a

ξ

¸ g(x)dx

a

b

g(x)dx, ξ

which was to be proved. Remark also that from the above it results also that the property is true also for f monotone increasing function, since we can apply the proved result for the function −f , which is decreasing. Therefore from (3) it results relation (2). 291

Before the demonstration of Bonnet’s formula, we will indicate some works where one can find particular cases or other names. In [1], Theorem 2 is proved for continuous f, g and increasing f ; and on page 341 it is stated that it holds true also for monotone f , and g as an integrable derivative. (Attention: a function which is a derivative, is not necessarily integrable!). In [2], p.337, the Theorem appears as (2), in which f (a) and f (b) are replaced with f (a + 0), resp. f (b − 0), with f, g bounded and integrable, f monotone. In [3], p.239, there is an incomplete proof. In [4], along with many interesting applications of the mean value theorems, appears also Bonnet’s formula, without a proof. In [7] relation (2) is called as ”Weierstrass formula”, while in [13] it is attributed to Du Bois-Reymond. In [8], the Bonnet formula is called as the ”second mean value theorem” with increasing f and integrable g. The proof given there is quite complicated. Finally, we mention [9], where the result (2) is proved for f, g continuous functions, f having a continuous derivative, and a constant sign. The book [6] (p.155) contains some interesting applications. See also [4], [5], [6], [11]-[14]. 2. The proof of Bonnet’s formula (Relation (3)) If f (a) = 0, then from 0 ≤ f (x) ≤ f (a) we get f (x) = 0, thus any ξ ∈ [a, b] is acceptable. Let us suppose thus f (a) > 0, and consider a division ∆ = {x0 , . . . , xn+1 } of [a, b] with x0 = a, xn+1 = b. Put S=

n X (xk+1 − xk )f (xk )g(xk ),

S0 =

n X

(xk+1 − xk )f (xk )λk ,

k=0

k=0

where λk ∈ [mk , Mk ], with mk , Mk being the extreme values of g on [xk , xk+1 ]. Suppose that λk is exactly the intermediate point which appears in the first mean value theorem (relation (1)) of g on [xk , xk+1 ]: Z xk+1 g(x)dx = (xk+1 − xk )λk . xk

Put

Z

x

G(x) =

g(t)dt. a

The function G being continuous, attains its margins m1 and M1 . From the equality G(xk+1 ) − G(xk ) = (xk+1 − xk )λk , we get 0

S =

n X

f (xk )[g(xk+1 ) − G(xk )] = −f (a)G(a) + G(x1 )[f (a) − f (x1 )] + · · · +

k=0

292

+G(xn )[f (xn−1 ) − f (xn )] + G(b)f (xn ). By G(a) = 0 and f decreasing, we get S 0 ≥ m1 [f (a) − f (x1 ) + · · · + f (xn−1 − f (xn ) + f (xn )] = m1 f (a), S 0 ≤ M1 [f (a) − f (x1 ) + · · · + f (xn−1 − f (xn ) + f (xn )] = M1 f (a) Let

Z

(4)

b

A=

f (x)g(x)dx. a

In what follows we shall prove that m1 f (a) ≤ A ≤ M1 f (a).

(5)

Indeed, if we would have e.g. A > M1 f (a), from the definition of Riemann’s integral and Darboux’s theorem (applied for the function g) we could write: ∀ ε > 0, ∃ δ > 0 such that k∆k < δ ⇒ |A − S| < ε and D =

n X

(xk+1 − xk )(Mk − mk ) < ε/f (a).

k=0

On the other hand, n X (xk+1 − xk )f (xk )|g(xk ) − λk | ≤ Df (a) < f (a)ε/f (a) = ε. |S − S | ≤ 0

k=0

Now, |A − S| < ε, |S − S 0 | < ε and A > M1 f (a), so 2 − ε < S 0 − A < 2ε. In other words S 0 → A > M1 f (a), thus for sufficiently large n we have S 0 > M1 f (a), a contradiction to (4). Let now α = A/f (a). On base of (5) we have m1 ≤ α ≤ M1 , and G being continuous, by the intermediate value property (i.e. ”Darboux property”) there is a ξ ∈ [a, b] with G(ξ) = α. This implies the equality Z b Z ξ f (x)g(x)dx = f (a) g(x)dx. a

a

See also [10] (where in Part II a proof based on Abel’s inequality is given). 3. Now, applying (2), and using a method by E.L. Stark [11], a rather elementary proof of Euler’s sum ∞ X 1 π2 = k2 6 k=1

293

(6)

will be given. The following identity is well known: n

Dn (x) =

1 X sin(2n + 1)x/2 + cos kx = 2 2 sin x/2

(x ∈ R, n ∈ N).

(7)

k=1

Set

Z Mn =

π

0

tDn (t)dt.

By employing the polynomial representation from (7), by partial integration we get ( ) m π2 X 1 M2m−1 = 2 − (m ∈ N). (8) 8 (2k − 1)2 k=1

On the other hand, by using the closed representation of (7) and applying x x the second mean value theorem (2) for f (x) = sin , x ∈ (0, π); f (0) = 1, 2 2 ³ π x´ f (π) = and g(x) = sin (4m − 1) we get 2 2 ¶ Z πµ x x/2 sin(4m − 1) dx M2m−1 = sin x/2 2 0 ½ ¾ ³π ´ ξ 1 =2 1+ − 1 cos(4m − 1) 2 2 4m − 1 µ ¶ 1 =O as m → ∞ (0 ≤ ξ ≤ π). (9) m By combining (8) and (9), we obtain ∞ X k=1

1 π2 , = (2k − 1)2 8

which immediately implies relation (6). 4. We note that the second mean value theorem of integral calculus (2) has many other applications in mathematics. For example in [5] it is given an application to the proof of Taylor’s formula. It is also important in the theory of trigonometric series (see e.g. [14]), or Laplace transforms (see e.g. [12]).

References [1] M. Balázs, J. Kolumbán, Mathematical analysis (Hungarian), Ed. Dacia, Cluj, 1978. 294

[2] G. Chilov, Analyse mathématique, Fonctions d’une variable, Ed. Mir, Moscou, 1978. [3] M. Craiu, M.N. Ro¸sculet¸, Problems of mathematical analysis (Romanian), Ed. Did. Ped., Bucure¸sti, 1976. [4] C.V. Cr˘aciun, Mean value theorems of mathematical analysis (Romanian), Univ. Bucure¸sti, 1986. [5] R.R. Goldberg, Methods of real analysis, Wiley, 1964. [6] N.M. G¨ unther, R.O. Cuzmin, Problems of higher mathematics (Romanian), vol.II, Ed. Tehnic˘ a, Bucure¸sti, 1950, 155-157. [7] C. Meghea, The basics of mathematical analysis (Romanian), Ed. S¸tiint¸. Enc., Bucure¸sti, 1977. [8] S.M. Nikolsky, A course of mathematical analysis, vol.I, Mir Publ., Moscou, 1977. [9] C. Popa, V. Hiri¸s, M. Megan, Introduction to mathematical analysis via exercises and problems (Romanian), Ed. Facla, 1976. [10] J. Sándor, On the second mean value theorem for integrals, Lucr. Semin. Didactica Mat., 5(1989), 275-280; II ibid., 2005, to appear. [11] E.L. Stark, Application of a mean value theorem for integrals to series summation, Amer. Math. Monthly, 1978, 481-483. [12] D. Widder, The Laplace transform, Princeton, 1941. [13] E.T. Whittaker, G.N. Watson, A course of modern analysis, Cambridge, 1969. [14] A. Zygmund, Trigonometric series, 2nd ed., vol.I, (p.58), Cambridge, 1959.

295

296

Chapter 7

Functional equations and inequalities ”... If I feel unhappy, I do mathematics to become happy. If I am happy, I do mathematics to keep happy.” (Alfred Rényi)

”... The authors have chosen to emphasize applications, though not at the expense of theory.” (J. Aczél and J. Dhombres)

297

1

The Bohr-Mollerup theorem

1. In 1922 H. Bohr and J. Mollerup [3] discovered the following surprising fact: Theorem 1. The only log-convex solution f : (0, ∞) → (0, ∞) to the functional equation f (x + 1) = xf (x), x > 0, satisfying f (1) = 1 is f (x) = Γ(x), the Euler Gamma function. Proof. This proof is based on ideas by E. Artin [1]. It is easy to see that Z ∞ Γ(x) = e−t tx−1 dt, x > 0 0

satisfies these conditions. Clearly Γ(1) = 1. By partial integration it follows that Γ(x + 1) = xΓ(x). Finally, by Hölder’s inequality one can prove that Γ is log-convex. 1 1 Indeed, let p > 1 and suppose that + = 1. Then p q ¶ Z ∞ µ Z ∞³ ´ 1 ³ y 1 t´ x y x y − t x −p − − −t p + q −1 + = e t Γ dt = e ptp e q q e q dt p q 0 0 µZ

∞

≤

−t x−1

e t

¶1/p µZ dt

0

∞

−t y−1

e t

¶1/q dt

0

by the Hölder inequality ¯Z ¯ ¯ ¯

0

∞

¯ µZ ¯ f (t)g(t)dt¯¯ ≤

∞

p

¶1/p µZ

|f (t)|

0

∞

q

¶1/q

|g(t)|

,

0

1 =λ∈ p (0, 1). Then Γ(λx + (1 − λ)y) ≤ (Γ(x))λ (Γ(y))1−λ , which shows that log Γ is a convex function on (0, ∞). We now prove that the given conditions uniquely determines the function f . Clearly, it is sufficient to consider x ∈ (0, 1]. Since f (1) = 1, we get f (2) = 1, f (3) = 1 · 2, . . . , f (n) = (n − 1)! (by mathematical induction) for any positive integer n ≥ 1. Let now n ≥ 2 be any integer and 0 < x ≤ 1. Since log f is convex, one has that x

applied to the functions f (t) = e−t/p t p

− p1

y

, g(t) = e−t/q t q

− 1q

. Let now

log f (−1 + n) − log f (n) log f (x + n) − log f (n) ≤ (−1 + n) − n (x + n) − n 298

≤ i.e.

log f (1 + n) − log f (n) (1 + n) − n

log f (x + n) − log(n − 1)! ≤ log n. x Writing these inequalities without logarithms, we get log(n − 1) ≤

(n − 1)x (n − 1)! ≤ f (x + n) ≤ nx (n − 1)!

(1)

Since f (x + 1) = xf (x), we have f (x + 2) = x(x + 1)f (x), . . . , f (x + n) = x(x + 1) . . . (x + n − 1)f (x), (1) can be written also as (n − 1)x (n − 1)! nx (n − 1)! ≤ f (x) ≤ . x(x + 1) . . . (x + n − 1) x(x + 1) . . . (x + n − 1)

(2)

By writing n + 1 in place of n in the left side of (2), we can write: nx n! nx n! x+n ≤ f (x) ≤ · , x(x + 1) . . . (x + n) x(x + 1) . . . (x + n) n or f (x)

n nx n! ≤ ≤ f (x). x+n x(x + 1) . . . (x + n)

(3)

(4)

Relation (4) shows that f (x) is uniquely determined by nx n! n→∞ x(x + 1) . . . (x + n)

f (x) = lim

(5)

This finishes the proof of the theorem, which shows also that one has the equality nx n! Γ(x) = lim (6) n→∞ x(x + 1) . . . (x + n) Remark. Equality (6) can be proved also by e.g. the Lebesgue (domination) criterion: Let (fn ) be a measurable sequence of functions on (a, b) and a.e. fn −→ f . Suppose there exists function g such that |fn (u)| ≤ g(u) for all u ∈ (a, b). Then f is measurable, and one has Z b Z b lim fn (u)du = f (u)du. n→∞ a

Z Let an (x) =

1

a

(1 − t)n tx−1 dt. By partial integration we get

0

n an (x) = x

Z

1

(1 − t)n−1 tx ,

0

299

and repeating the argument n times, we get an (x) = By letting t =

u , we obtain n 1 an (x) = x n

i.e.

n! . x(x + 1) . . . (x + n)

Z

n³

1− 0

n!xn = x(x + 1) . . . (x + n) Put now

³ Since 1 −

Z

u ń x−1 u du, n

n³

1− 0

u ń x−1 u du n

(∗)

( ³ u ń x−1 u , 0n g(u) = e−u ux−1 ,

u ń n

u ∈ (a, b) = (0, +∞).

< e−u , by the above criterion by Lebesgue Z

lim

n³

n→∞ 0

1−

u ń x−1 u du = n

Z

∞

e−u ux−1 du,

0

and by (∗) the result follows. 2. Let 0 < q < 1. In 1869 J. Thomae [5], and independently in 1904 F.H. Jackson [4] introduced the so-called q-Gamma function: Γq : (0, ∞) → (0, ∞) by (1 − q)(1 − q 2 ) . . . (1 − q n+1 ) , n→∞ (1 − q x )(1 − q x+1 ) . . . (1 − q x+n )

Γq (x) = (1 − q)1−x lim

x > 0.

(7)

In 1978 R. Askey [2] proved the following: Theorem 2. The only log-convex solution f : (0, ∞) → (0, ∞) to the functional equation f (x + 1) =

1 − qx f (x), 1−q

x>0

satisfying f (1) = 1 is f (x) = Γq (x). In 1997 R. Webster [6] obtained a common generalization to the above theorems. In what follows we will obtain a simplified proof. 300

Theorem 3. Let g : (0, ∞) → (0, ∞) be a log-concave function which has the property that for all w > 0 one has g(x + w)/g(x) → 1 as x → ∞. Then the only log-convex solution to the functional equation f (x + 1) = g(x)f (x),

x > 0,

satisfying f (1) = 1 is g(n) . . . g(1)(g(n))x . n→∞ g(n + x) . . . g(x)

f (x) = lim

Proof. The uniqueness of f , i.e. equality (8), follows on the same lines as in the proof of Theorem 1, by using the log-convexity of f and the property g(x + w)/g(x) → 1 as x → ∞ for all w > 0. We must prove also the existence of such a solution. For each n ∈ N define a function fn : (0, ∞) → (0, ∞) by fn (x) = Then

g(n) . . . g(1)(g(n))x , g(n + x) . . . g(x)

x > 0.

(9)

 (g(n + 1))x+1   f (x) = fn (x)    n+1 g(n + x + 1)(g(n))x      fn+1 (x + 1) =

(10) g(n) g(x)fn (x). g(n + x + 1)

Let 0 < x ≤ 1. By the log-concavity of g one can write (g(n + 1))x+1 ≥ (g(n))x g(n + x + 1), which by the first relation of (10) shows that fn+1 (x) ≥ fn (x), i.e. the sequence (fn (x))n is decreasing. Writing n times the log-concavity of g, we get: g(x + 1) ≥ (g(1))1−x (g(2))x , . . . , g(x + n) ≥ (g(n))1−x (g(n + 1))x , and their product, together with (9) shows that µ ¶x (g(1))x g(n) fn (x) ≤ g(x) g(n + 1) Now, it can be shown that g(n) ≤ g(n + 1), so (11) implies µ ¶ g(1) x fn (x) ≤ , g(x) 301

(11)

i.e. (fn (x))n is bounded above. Thus (fn (x))n being bounded, and monotone, converges to a function fe : (0, 1] → (0, ∞) by g(n) . . . g(1)(g(n))x fe(x) = lim for 0 < x ≤ 1. n→∞ g(n + x) . . . g(x)

(12)

By (10) it follows that (12) may be extended to all x > 0 defining a function f : (0, ∞) → (0, ∞) by g(n) . . . g(1)(g(n))x , n→∞ g(n + x) . . . g(x)

f (x) = lim

x > 0.

We have to prove only that g(n) ≤ g(n + 1), or more generally that g is an increasing function. Indeed, let 0 < a < b. Since log g is concave on (0, ∞), log g(a + x) − log g(a) ≥ log g(b + x) − log g(b),

x > 0,

so

g(a) g(a + x) ≤ . g(b) g(b + x) g(a + x) → 1, thus implying g(a)/g(b) ≤ Now, by assumption, as x → ∞, g(b + x) 1. Since a < b, this proves that g is an increasing function. 1 − qx Remarks. g(x) = x and g(x) = (0 < q < 1) satisfy the conditions 1−q of Theorem 3.

References [1] E. Artin, Einf¨ uhrung in die Theorie der Gammafunktion, Teubner, Leipzig, 1931. [2] R. Askey, The q-gamma and q-beta functions, Appl. Anal. 8(1978), 125141. [3] H. Bohr and J. Mollerup, Laerebog i matematisk Analyse, vol.III, 149164. [4] F.H. Jackson, A generalization of the functions Γ(n) and xn , Proc. Roy. Soc. London, 74(1904), 64-72. [5] J. Thomae, Beitr¨ age zur Theorie der durch der Heinesche Reihe..., J. Reine Angew. Math. 70(1869), 258-281. [6] R. Webster, Log-convex solutions to the functional equation f (x + 1) = g(x)f (x): Γ-type functions, J. Math. Anal. Appl. 209(1997), 605-623. 302

2

On certain functional equations containing more unknown functions

1. Problem C:564 proposed by V. B˘andil˘ a [2] asks for the determination of three continuous functions f, g, h : R → (0, ∞) such that µ ¶ x+y f (x)g(y) = h , x, y ∈ R. (1) 2 This problem can be reduced, as we shall see, to the solution of Cauchy’s functional equation F (x + y) = F (x) + F (y),

x, y ∈ R.

(2)

On the other hand, we will show that, equation (1) is a particular case of a ”Popoviciu type equation”, studied and generalized by Stamate [3], Ianić and Peˇcarić [4], and Acu [5]. See also [6]. In the second part, we shall study a functional equation suggested by the equation by Popoviciu. As a particular case, we will obtain the solutions to a problem by Vlada [7]. Then we will suggest some generalizations which are analogous to the ones studied by Stamate. 2. All continuous solutions of equation (2) are of the form F (x) = ax,

x ∈ R,

where a is fixed. Indeed, by letting successively y = x, y = 2x, . . . we get F (2x) = 2F (x), 1 F (3x) = 3F (x), . . . , F (nx) = nF (x) (by mathematical induction). For x = n µ ¶ µ ¶ ³m´ F (1) 1 m 1 1 = = a. Similarly, F = a. From the we get F = mF n n n n n n observation F (0) = 0 one gets F (−x) = −F (x), so we get finally F (r) = ar for all rational numbers r. Let now x ∈ R \ Q, and consider a sequence (xn ) of rational numbers with xn → x as n → ∞. By F (xn ) = axn , and the continuity of F we can deduce F (x) = ax (where in fact, a = F (1)). 3. We now consider equation (1). Letting y = x, it results h(x) = f (x)g(x). So h(y) = f (y)g(y), and by substitution, µ ¶ x+y h(x)h(y) = h2 . 2 The function h being strictly positive, we can define H by h(x) = eH(x) . This yields the equation (known also as ”Jensen’s functional equation”) µ ¶ x+y H(x) + H(y) = 2H . 2 303

³x´ ³x´ For y = 0 we obtain H(x) = 2H − H(0), thus 2H − H(0) + 2 2 µ ¶ ³y ´ ³x´ x+y − H(0) = 2H − H(0). 2H . Let us introduce h1 (x) = H 2 2 2 Then we easily arrive at Cauchy’s functional equation h1 (x + y) = h1 (x) + h1 (y). By 2., the most general continuous solutions to this equation are h1 (x) = ax, thus H(x) = 2ax + b (where b = H(0)). Therefore h(x) + e2ax+b = cekx (where c > 0, k ∈ R). By applying (1) to y = 0, we get finally: f (x) = h2

³x´ 2

/g(0) =

c2 kx e , g(0)

g(y) =

c2 ky e , f (0)

where c2 = f (0)g(0). 4. A more general equation like (1) is h(x + y) = F (x)G(y),

(3)

where h, F, G : R → R are continuous functions. Indeed, f (2x) = F (x), g(2y) = G(y) imply to the above equation, with the assumption that F (x), G(x) ∈ (0, ∞). This equation is a particular case of a ”Popoviciu type equation” (see [5]): h(x + y + z) = F (x)G(y, z),

(4)

where G is a continuous function of two variables. Let us remark that, for z = 0, and with the notation G(y, 0) = G(y), from (4) we arrive at (3). It is possible to prove that the solutions of (4) are given by: h(x) = abecx , F (x) = aecx , G(y, z) = bec(y+z) where a, b, c are certain constants. From this we are able to reobtain the solutions obtained in 3. 5. Let us consider in what follows the functional equation f (x + y + z) = g(x) + h(y, z)

(5)

suggested by the equation (4), of Popoviciu type. We will obtain all continuous solutions f, g : R → R, h : R2 → R. Putting x = 0, we get f (y + z) = g(0) + h(y, z); thus f (x + y + z) = g(x) + f (y + z) − g(0). 304

Applying (5) for y = z = 0, it results f (x) = g(x) + h(0, 0) = g(x) + f (0) − g(0). Thus, by substitution, f (x + y + z) = f (x) − f (0) + g(0) + f (y + z) − g(0) = f (x) + f (y + z) − f (0). For x = u, y + z = v and the notation f (x) − f (0) = f1 (x), we get f1 (u + v) = f1 (u) + f1 (v), i.e. the Cauchy functional equation (2). Since f1 (x) = Ax, we get f (x) = f (0) + Ax = Ax + B. The function g can be found by g(x) = Ax + B − (f (0) − g(0)) = Ax + C; and similarly h by h(y, z) = f (y + z) − g(0) = A(y + z) + D (A, B, C, D ∈ R). Remark. This functional equation contains the special case h(y, z) = h(y) + k(z) (see [7]). From A(y + z) + D = h(y) + k(z) with z = 0 it immediately results that h(y) = Ay + A1 , k(z) = Ax + A2 (for y = 0). 6. From the generalizations of the functional equation (5) we may consider the followings: 1) f (x1 + x2 + · · · + xn ) = f1 (x1 ) + f2 (x2 , x3 , . . . , xn ), where f1 : R → R, f : R → R, f2 : Rn−1 → R are continuous functions; 2) f (x1 +x2 +· · ·+xn ) = f1 (x1 )+f2 (x2 )+· · ·+fn−2 (xn−2 )+fn−1 (xn−1 xn ), where f, f1 , . . . , fn−2 : R → R, fn−1 : R2 → R are continuous. These equations can be studied with the methods shown before. For example, for eq. 1) we proceed as follows: for x1 = 0 we get f (x2 + · · · + xn ) = f1 (0) + f2 (x2 , x3 , . . . , xn ), thus f (x1 + · · · + xn ) = f1 (x1 ) + f (x2 + · · · + xn ) − f1 (0). For x2 = · · · = xn = 0 we get f1 (x1 ) = f (x1 ) − f (0) − f1 (0), thus by substitution, f (x1 + · · · + xn ) = f (x1 ) + f (x2 + · · · + xn ) − k (k = constant). Put F (x) = f (x) − k, x1 = u, x2 + · · · + xn = v. Then we get F (u + v) = F (u) + F (v), etc. 7. As a final remark, we notice that the above equations may be further generalized, obtaining the so-called ”Pexider type” equations (see [1]). 305

References [1] J. Aczél, Lectures on functional equations and their applications, Academic Press, 1966. [2] V. B˘andil˘a, Problem C:564, Gazeta Mat. Seria B, 2/1986. [3] I. Stamate, Equations fonctionnelles contenant plusieurs fonctions inconnues, Univ. Beograd Publ. Elektr. Fak. Ser. Mat. Fiz., nr.354-356, 1971. [4] R. Ianić, J.E. Peˇcarić, On some functional equations for functions of several variables, Ibid., nr.716-734, 1981. [5] D. Acu, On a functional equation of Popoviciu type (Romanian), Gaz. Mat. Seria A, 3-4/1984. [6] J. Sándor, On a functional equation (Romanian), Lucr. Sem. Did. Mat., 6(1990), 293-296. [7] M. Vlada, Problem 13259, Gaz. Mat. 6/1973.

3

Locally integrable solutions of Cauchy’s functional equation

A function f : R → R is called locally integrable, if it is integrable over every finite interval of R. Theorem. The locally integrable solutions of the equation f (x + y) = f (x) + f (y),

x, y ∈ R

(1)

are f (x) = cx, c = constant. Proof. On the basis of (1), and the hypothesis of local integrability one easily verifies the following identity: Z x+y Z x Z y yf (x) = f (t)dt − f (t)dt − f (t)dt. (2) 0

0

0

Relation (2) implies at once xf (y) = yf (x),

x, y ∈ R.

Letting x = x0 6= 0, from (3) we get f (y) = cy, where c = finishes the proof of the theorem. 306

(3) f (x0 ) ; and this x0

Remark. W. Sierpinski [1] proved that the above result remains valid under the weaker hypothesis that f (x) is measurable. The given simple proof of the above theorem is due to H.N. Shapiro [2].

References [1] W. Sierpinski, Sur l’équation fonctionnelle f (x + y) = f (x) + f (y), Fundamenta Math., 1(1920), 116-122. [2] H.N. Shapiro, A micronote on a functional equation, American Math. Monthly, 80(1973), no.9, 1041.

4

Generalizations of Haruki’s and Cior˘ anescu’s functional equations 1. In 1979 S. Haruki [5] proved that the functional equation f (x) − g(y) = h(x + y), x−y

x 6= y,

(1)

where f, g, h : R → R, has f (x) = g(x) = ax2 + bx + c and h(x) = ax + b as its only solutions, without any regularity condition. His proof went by reduction to the well known Jensen functional equation (see e.g. [1]). In 1985 J. Aczél [2] used an interesting elementary argument to prove the above considered result. In what follows we shall apply Aczél’s method in order to solve some functional equations, one of which generalizes Haruki’s equation, while the others extend certain results by N. Cior˘anescu [3] (see also [5]). 2. Let g : R → ∞R be a given additive function, and let us consider the equation f (x) − F (y) = h(x + y), g(x) 6= g(y), (2) g(x) − g(y) where f, F, h : R → R are the unknown functions. Theorem 1. Without any regularity conditions, all solutions of equation (2) are given by ³x´ f (x) = F (x) = ag 2 (x) + bg(x) + c, h(x) = ag + b, 2 where a, A, b, c are certain real constants. Proof. By interchanging x and y we obtain easily f = F , consequently the equation becomes f (x) − f (y) = (g(x) − g(y))h(x + y). 307

(3)

If f satisfies this equation, so does f + b (b = constant), so we may suppose that f (0) = 0. Put y = 0 in (3) in order to get f (x) = g(x)h(x)

(4)

(because of g(0) = 0, g being an additive function, i.e. g(x + y) = g(x) + g(y) for all x, y ∈ R). This transforms (3) into g(x)h(x) − g(y)h(y) = [g(x) − g(y)]h(x + y).

(5)

Again, we may suppose h(0) = 0. Therefore, putting x = −y into (5), we get g(−y)h(−y) = g(y)h(y).

(6)

We take this into consideration when replacing y by −y in (5) getting [g(x) − g(y)]h(x + y) = [g(x) − g(−y)]g(x − y). Substituting here x + y = u, x − y = v, we have: µ µ ¶ µ ¶¶ µ µ ¶ µ ¶¶ u+v u−v u+v v−u g −g h(u) = g −g h(v). 2 2 2 2

(7)

(8)

The function g being additive, one has: ¶ µ ¶ µ ³v ´ u−v u+v −g = 2g g , 2 2 2 and similarly

µ g

u+v 2

so one arrives to g

¶

³v ´

µ −g

v−u 2

= 2g

³u´

³u´

2

,

h(v). (9) ³u´ Let v = v0 in (9). Thus we get h(u) = ag . If we do not assume 2 ³u´ + b. By (4) this gives h(0) = 0, we have in general h(u) = ag 2 ³ ³x´ ´ f (x) = g(x) ag + b = Ag 2 (x) + bg(x); 2 2

h(u) = g

¶

2

and if we do not assume f (0) = 0, we have: f (x) = Ag 2 (x) + bg(x) + c. 308

3. N. Cior˘anescu [3] considered the following equations: f (x) − f (y) 1 = [f 0 (x) + f 0 (y)]; x−y 2 µ

f (x) − f (y) x−y

¶2

= f 0 (x)f 0 (y),

(10)

(11)

where f is differentiable. Equations (10) and (11) were introduced in connection with some properties of conics. In what follows we will study the more general equations f (v) − f (u) = h(u) + h(v), u 6= v, v−u ¶ µ f (v) − f (u) 2 = h(u)h(v), u = 6 v, v−u 1 f (v) − f (u) = , v−u h(u) + h(v) Equation (14) arises from

u 6= v.

(12) (13) (14)

f (v) − f (u) 2f 0 (u)f 0 (v) = 0 , not studied by v−u f (u) + f 0 (v)

Cior˘anescu. In what follows, we shall apply the method to equations (12) and (14). Theorem 2. Let f, h : R → R satisfy equation (12). If h ≡ 0, then f = constant. If h 6≡ 0, then h(u) = au + b, and satisfies f (u) = au2 + b1 u + c, where a, b1 , c are constants. Proof. Assume as above f (0) = 0. Put u = 0 in order to deduce f (v) = v[h(v) + h(0)],

v 6= 0.

(15)

Therefore, we have vh(v) − uh(u) + vh(0) − uh(0) = (h(u) + h(v))(v − u).

(16)

This is equivalent to vh(u) − uh(v) = (v − u)h(0).

(17)

Let v = v0 (6= 0) in (17). This implies h(u) =

1 uh(v0 ) − h(0) [uh(v0 ) + vh(0) − uh(0)] = + h(0). v0 v0 309

(18)

We have to distinguish here two cases: Case 1. h ≡ 0. Then (15) gives f ≡ 0. if we do not assume f (0) = 0, we get f = constant. Case 2. h 6≡ 0, i.e. there exists v0 ∈ R such that h(v0 ) 6= 0. Then obviously v0 6= 0 (see (18)), so we can write: h(u) = au + b, implying f (u) = au2 + b1 u. Finally, it results f (u) = au2 + b1 u + c. Remark. As an application we can find immediately the solutions of the Cior˘anescu equation (10): One finds f (x) = Ax or f (x) = Ax2 + Bx + C. Theorem 3. Suppose that f, h : R → R satisfy the equation (14), where h(u) + h(v) 6= 0 for u 6= v. If h = constant, then f (u) = Au + B (A, B constants). If h 6= constant, then h2 (u) = au + b (a 6= 0), and u + f (0). f (u) = A + h(u) Proof. Assuming f (0) = 0, and letting u = 0, we get 1 f (v) = , v h(0) + h(v) so that we obtain µ

u v − a + h(v) a + h(u)

v 6= 0

(19)

(h(u) + h(v)) = v − u.

(20)

¶

Here h(u) 6= −a, with a = h(0) for all u ∈ R. After some elementary operations, (20) becomes equivalent to vh2 (u) − uh2 (v) = a2 v − a2 u. Letting v = v0 in (21), where v0 6= 0, we obtain: · 2 ¸ h (v) − h2 (0) + h2 (0). h2 (u) = u v0 Case 1. h(u) = h(0) (since h(u) 6= −h(0) by assumption). This gives f (u) =

u u + f (0) = , a + h(u) 2a

and if we do not assume f (0) = 0, then f (u) =

u + f (0). 2a 310

(21)

(22)

Case 2. h(u) 6= h(0) for some u 6= 0, i.e. h(u0 ) 6= h(0) for some u0 6= 0. Then we can easily deduce h2 (u) = Au + B (A 6= 0), B = a2 , and f (u) =

u + f (0), a + h(u)

where in fact A = (h2 (u0 ) − h2 (0))/u0 .

References [1] J. Aczél, Lectures in functional equations and their applications, Academic Press, 1966. [2] J. Aczél, A mean value property of the derivative of quadratic polynomials - without mean values and derivatives, Math. Mag. 586(1985), 42-45. [3] N. Cior˘anescu, Deux équations fonctionnelles et quelques propriétés de ´ certaines coniques, Bull. Math. Phys. Pures Appl. Ecole Polyt. Bucarest, 9(1937-1938), 52-53. [4] S. Haruki, A property of quadratic polynomials, Amer. Math. Monthly, 86(1979), 577-579. [5] J. Sándor, On certain functional equations, Itinerant Sem. Funct. Eq. Approx. Convexity 1988, Cluj, 285-288.

5

The equation Ã nfunctional ! n X X f xk = (f (xk ))k k=1

k=1

In this note we obtain a solution of OQ.32 (Octogon Math. Mag., 4(1996), no.1, p.85) proposed by M. Bencze and F. Popovici. This problem asks for the determination of all functions f : R → R having the property f (x1 + · · · + xn ) = f (x1 ) + f 2 (x2 ) + · · · + f n (xn ),

(1)

for all x1 , . . . , xn ∈ R (n fixed). Here f n (x) := (f (x))n . When f is continuous, a more general equation has been studied by the author [2]. For functional equations without any regularity we quote [3]. Certain (quite difficult) open problems were stated in [4]. 311

1) First assume that n ≥ 4. We shall prove that equation (1) is equivalent with the following system ½ 2 f (x) = f (x), x∈R (2) f (x + y) = f (x) + f (y), x, y ∈ R. By putting x1 = x2 = · · · = xn = 0 in (1) we get f 2 (0)[1 + f (0) + · · · + f n−2 (0)] = 0.

(3)

Since the equation 1+z+· · ·+z n−2 = 0 has only complex roots for n−2 ≥ 2, from (3) we obtain that f (0) = 0. Letting x1 = x3 = · · · = xn = 0, x2 = x in (1) we can deduce that f 2 (x) = f (x). Similarly f 3 (x) = · · · = f n (x) = f (x). Thus f (x1 + · · · + xn ) = f (x1 ) + · · · + f (xn ), and taking x1 = x, x2 = y, x3 = 0, . . . , xn = 0 by f (0) = 0 we get the second equation of (2). Reciprocally, each solution of the system (2) is a solution of equation (1). This follows easily by induction. Now, if we admit that f satisfies a regularity condition (for example, f is continuous at a point, is monotone, or is measureable, etc.), then it is well-known that Cauchy’s function equation (i.e. the second equation of (2)) has the general solution f (x) = cx, c ∈ R (fixed). Since c2 x2 = cx, ∀ x ∈ R, clearly c = 0 (e.g. x = 1 and x = 2). Thus the general solution of equation (1) is f ≡ 0 (obviously, if one of the above stated regularity conditions is satisfied). Remark 1. For the above stated results on Cauchy’s functional equation, see [1]. 2) For n = 1 the equation (1) becomes f (x1 ) = f (x1 ), x1 ∈ R, and all functions f : R → R satisfy this property. 3) Let n = 2. Then, as in (3), we get f (0) = 0, and we obtain the same solutions as in case 1). 4) Let n = 3. Then the equation becomes f (x + y + z) = f (x) + f 2 (y) + f 3 (z),

x, y, z ∈ R.

(4)

From x = y = z = 0 we have f (0)[1 + f (0)] = 0. Thus, we have to consider two cases: a) f (0) = 0, b) f (0) = −1. In case a) we get, if we put x = z = 0 that f (y) = f 2 (y) and the same solutions as in 1) are obtained. When f (0) = −1, by substituting x = z = 0, we can deduce that f (y) = f 2 (y) − 2. (5) Let y = 0 in (4). Then f (x + z) = f (x) + 1 + f 3 (z). Now if x = 0 in this equation, clearly f (z) = f 3 (z). On the other hand, from (5) we have 312

f 3 (z) = (f (z) + 2)f (z) = f 2 (z) + 2f (z) = 3f (z) + 2. From 3f (z) + 2 = f (z) we obtain that f (z) = −1, z ∈ R. This is a most general solution of equation (4), when f (0) = −1. Remark 2. The discontinuous general solutions of Cauchy’s functional equation can be obtained via the so called ”Hamel basis”. This is in fact a base of Q-vectorial space R. The interested reader can deduce that the most general solution of system (2) if f ≡ 0. Thus, this is the general solution of equation (1) in case n ≥ 4 and n = 2 or n = 3, f (0) = 0.

References [1] J. Aczél, Lectures on functional equations and their applications, Academic Press, New York, San Francisco, London, 1966. [2] J. Sándor, Asupra unei ecuat¸ii funct¸ionale, Lucr. Semin. Did. Mat., 6(1990), 293-296. [3] J. Sándor, On certain functional equations, Itinerant Seminar Funct. Eq. Approx. and Convexity, Babe¸s-Bolyai Univ., 1988, 285-288. [4] J. Sándor, On certain open problems in the theory of functional equations, Proc. Symp. Appl. of Funct. Eq. in Ed., Science and Prod., 4th iune 1988, Odorheiu Secuiesc, Romania.

6

Two functional equations

Problem OQ.57 (M. Bencze and F. Popovici) from Octogon Math. Mag., 4(1996), no.2, p.78 asks for the determination of positive functions f with the property ! Ã n n q X X f xk = f (xk ), ∀ x ∈ R (n fixed). k=1

k=1

The misprint in f (xk ) can be interpreted in two ways, namely, as f (xk ) or as f (xkk ). These two situations lead to two distinct functional equations; and our aim is to study both cases in this note. 1) The functional equation Ã n ! n p X X k f xk = f (xk ). (1) k=1

k=1

313

In this case, the method is very similar to that shown inpour Note [1]. Namely, let x1 = · · · = xn = 0 in (1). We get f (0) = f (0) + f (0)p + ··· + p n f (0), p and this implies f (0) = 0 (since on the right-hand side of 0 = f (0)+ · · ·+ n f (0) are only nonnegative members. Letting x3 = · · · = xn = 0, x1 = x, x2 = y we get f (x p+ y) = f (x) + f (y). But from x1 = 0, x3 = 0, . . . , xnp= 0 we obtain f (x2 ) = f (x2 ), and in the same manner, we have f (xn ) = n f (xn ). Thus f (x) = f 2 (x) = · · · = f n (x) (where f n (x) := (f (x))n ). Thus we get the system ½ f (x + y) = f (x) + f (y), x, y ∈ R, (2) f (x) = f 2 (x), x ∈ R. Reciprocally, each solution of system (2) is a solution of (1). We have to solve the system (2) for arbitrary nonnegative functions. This can be made by using the theory of Cauchy’s equation (i.e. the first equation of (2)) but we shall follow here a very simple other argument. Clearly one has f 2 (x + y) = f 2 (x) + f 2 (y) + 2f (x)f (y) or f (x + y) = f (x) + f (y) + 2f (x)f (y). This gives f (x)f (y) = 0, yielding f (x) = 0, ∀ x ∈ R. This is the most general solution of the functional equation (1). 2) The functional equation ! Ã n n q X X k xk = f f (xkk ). (3) k=1

k=1

By putting x1 = · · · = xn =p0 in (3) we obtain f (0) = 0. From x1 = x3 = · · · = xn = 0, we get f (x2 ) = f (x22 ), i.e. f (x2 ) = (f (x))2 for all x ∈ R. In the same manner we can deduce that (f (x))2 = f (x2 ), (f (x))3 = f (x3 ), . . . , (f (x))n = f (xn ).

(4)

We will show that the functional equation (3) is equivalent to the system f (x + y) = f (x) + f (y), f (x2 ) = (f (x))2 .

(5)

By f (x2 + y 2 ) = f (x2 ) + f (y 2 ) = f 2 (x) + f 2 (y) and f ((x + y)2 ) = f (x2 + 2xy + y 2 ) = f (x2 + y 2 ) + f (2xy) = (f (x))2 + (f (y))2 + f (2xy) = f 2 (x + y) = f 2 (x) + f 2 (y) + 2f (x)f (y) we get f (2xy) = 2f (x)f (y). 314

(6)

We shall prove that f (xy) = f (x)f (y). (7) µ ¶ 1 1 Indeed, let y = in (6). Then f (x) = 2f f (x) and if we suppose that 2 2 µ ¶ 1 = 0, then f (0) = 0, ∀ x ∈ R, and then clearly (7) is satisfied. If one f 2 µ ¶ µ ¶ 1 1 1 admits that f 6= 0, then clearly f = and from the first equation 2 2 2 of (5) we get f (1) = 1. Letting y = 1 in (6), we can deduce f (2x) = 2f (x), thus f (2xy) = 2f (xy). This relation combined with (6) gives the equation (7). Now, from (7) easily follow all relations (4). Thus we have proved that equation (3) is equivalent with the system (5), or, on base of (7), with the system ½ f (x + y) = f (x) + f (y) (8) f (xy) = f (x)f (y) By f (1) = f 2 (1) we have two possibilities: a) f (1) = 0, in which case f (x) = 0 is the most general solution of (8); and b) f (1) = 1. Since f (0) = 0, (8) represents nothing else than all automorphisms of the field of real numbers R. When f is continuous at a point, is monotonic, or measurable, etc., all solutions of (8) are f (x) = x, ∀ x ∈ R, and there are no other solutions.

7

A functional equation satisfied by the sin and identity functions Let f : R → R such that µ

(n + 1)x f (nx)f n X 2 ³x´ f (kx) = f k=1 2

¶ for all x ∈ R and all n ∈ N.

(1)

Clearly, f (x) = x and f (x) = sin x are examples. By writing (1) for n + 1 and subtracting, one obtains the equation ·µ ¶ ¸ n+1 ¶ f x · µ ³ nx ´¸ (n + 2)x 2 ³ ´ f [(n + 1)x] = −f , (2) f x 2 2 f 2 315

∀ x ∈ R, ∀ n ∈ N. In fact, (1) and (2) are equivalent. Indeed, by writing (2) for n = 1, 2, . . . , (n − 1), after addition one gets: µ ¶ µ ¶ 3x 3x f (x)f f f (2x) 2 2 ³x´ ³x´ f (2x) + f (3x) + · · · + f (nx) = − f (x) + f f 2 2 µ ¶ µ ¶ µ ¶ µ ¶ 3x 3x 5x 3x f (x) f (2x)f f (2x)f f f (3x) f 2 2 2 2 ³x´ ³x´ ³x´ ³x´ − + − + − ···+ f f f f 2 2 2 2 ¶ ³ ´ ¶ µ ¶ µ µ nx (n − 2)x (n − 1)x (n − 1)x f f f f 2 2 2 2 ³x´ ³x´ + − f f 2 2 ¶ µ ³ nx ´ ³ nx ´ µ (n − 1)x ¶ (n + 1)x f f f f 2 2 2 2 ³x´ ³x´ + − , f f 2 2 giving (1). Therefore we have to consider equation (2). Without any auxiliary assumption on f it seems difficult to deduce any significant result on f . Let us assume that f (0) = 0 and that there exists a function g : R → R such that ¶ µ ¶ µ a+b a−b g , ∀ a, b ∈ R. f (a) − f (b) = 2f 2 2 (For example f (x) = sin x and f (x) = x are such functions, with the selections g(x) = cos x or g(x) = 1). Then, by putting b = 0 one gets ³a´ ³a´ f (a) = 2f g . 2 2 Since µ ¶ ³ nx ´ ³ x ´ µ (n + 1)x ¶ (n + 2)x f −f = 2f g , 2 2 2 2 ³a´ ³a´ by the above equality f (a) = 2f g applied to a := (n + 1)x, we can 2 2 deduce that (2) holds true. All in all, we have proved that all functions f with the property µ ¶ µ ¶ a−b a+b f (0) = 0, f (a) − f (b) = 2f g , a, b ∈ R 2 2 316

satisfy relation (1).

8

On certain functional equations by Rassias 1. Solve the equation: µ µ ¶ ¶ µ ¶ x x + f (xy) = f f + f (y) . f y y µ ¶ x Let f + f (x, y) = t. Then (1) can be written as y f (t) = t.

(1)

(2)

This means that if for each t ∈ A there exist x, y ∈ R, y 6= 0 such that µ ¶ x f + f (x, y) = t (3) y then the general solution of equation (1) is the identity function. For example, x when f : A → B is surjective (A ⊆ R such that x, y ∈ A ⇒ xy, ∈ A) then y (3) is satisfied. Indeed, let y = 1. Then 2f (x) = f

³x´ 1

t + f (x · 1) = t ⇔ f (x) = . 2

t t Let ∈ B. Since f is surjective, there exists x ∈ A with f (x) = . 2 2 2. Solve the equation: √ √ √ f (x xy + y x + y y) = f (f (x) + f (y)).

(4)

Put x → x2 , y → y 2 in (4) (where x, y ≥ 0). Then we get f (x3 y + xy 2 + y 3 ) = f [f (x2 ) + f (y 2 )].

(5)

Let x = 0 in (5). Then f (y 3 ) = f [f (0) + f (y 2 )]. For y = 0 in (5), we get f (0) = f [f (x2 ) + f (0)]. Since f [f (0) + f (x2 )] = f (x3 ), (put y → x in the above equality), we obtain f (0) = f (x3 ). (6) √ By putting x → 3 x, this implies f (x) = f (0) = constant. 317

3. Find the general solution of the equation f (xf (y) + yf (x) −

√ √ xf (y) − yf (x)) = f (x) + f (y).

(7)

Clearly x, y ≥ 0 in (1). Put y = 0 in (7). Then f ((x −

√ x)f (0)) = f (x) + f (0)

(8)

√ which for x = 0 gives f (0) = 0. Therefore (x − x)f (0) = 0, ∀ x ∈ A (A ⊂ [0, +∞)) implying f (0) = f (x) + f (0), i.e. 0 = f (x), ∀ x ∈ A. Reciprocally, f (x) = 0, ∀ x ∈ A satisfies (7). 4. What is the general solution of the equation: f (x)f (y) + f (y)f (x) = xf (x) + y f (x) .

(9)

We must have x, y > 0, f (x) > 0, f (y) > 0. Let x = 1 in (9). Then f (1)f (y) + f (y)f (1) = 1 + y f (1) .

(10)

Putting y = 1 in (10), we deduce 2f (1)f (1) = 2, i.e. f (1)f (1) = 1. Therefore f (1) = 1 and (10) implies f (y) = y. (11) By concluding, f (x) = x, ∀ x > 0. 5. We must solve the functional equation: f (xy) + f (x + y) = f (xy + x) + f (x − y).

(12)

Let f : A → R (A ⊆ R). Since x − y ∈ A, clearly 0 ∈ A. Put y = 0 in (12). Then we get f (0) + f (x) = f (x) + f (x), giving f (x) = f (0), ∀ x ∈ A. Therefore, f = constant. Reciprocally, f = constant satisfies equation (12).

References [1] Th.M. Rassias, OQ.738, Octogon Math. Mag., 9(2001), no.1, 1086. [2] Th.M. Rassias, OQ.739, Octogon Math. Mag., 9(2001), no.2, 1086. [3] Th.M. Rassias, OQ.735, Octogon Math. Mag., 9(2001), no.2, 1085. [4] Th.M. Rassias, OQ.736, Octogon Math. Mag., 9(2001), no.2, 1085. [5] Th.M. Rassias, OQ.737, Octogon Math. Mag., 9(2001), no.2, 1085. 318

9

Bartha’s functional equation

Let n ≥ 1 be fixed. We have to determine all functions f : R → R such that f 2 (x1 + x2 + · · · + xn ) = f 3 (x1 ) + f 3 (x2 ) + · · · + f 3 (xn ) (1) for all xk ∈ R (k = 1, n). Let n = 1. Then (1) gives f 2 (x1 ) = f 3 (x1 ), ∀ x1 ∈ R, i.e. f 2 (x1 )[f (x1 )−1] = 0. Let A = {x ∈ R : f (x) = 0}, B = {x ∈ R : f (x) = 1}. Then any function with the property A ∪ B = R satisfies the equality. Suppose n ≥ 2. By putting x1 = x2 = · · · = xn = 0, we get f 2 (0) = nf 3 (0), i.e. f 2 (0)[nf (0) − 1] = 0. i) f (0) = 0. By putting x2 = · · · = xn = 0 in (1) we get f 2 (x1 ) = f 3 (x1 ), ∀ x1 ∈ R, therefore f 2 (x1 + · · · + xn ) = f 2 (x1 ) + · · · + f 2 (xn ). Let g(x) = f 2 (x). Then g(x1 + · · · + xn ) = g(x1 ) + · · · + g(xn ) and letting x3 = · · · = xn = 0 we obtain g(x1 + x2 ) = g(x1 ) + g(x2 ).

(1)

Let x2 = −x1 in (1). Then 0 = g(x1 ) + g(x2 ), i.e. g(−x1 ) = −g(x1 ) giving 0 ≤ g(x1 ) = −g(−x1 ) ≤ 0 since g(x) = f 2 (x) ≥ 0 for all x ∈ R (particularly for x := x1 and x := −x1 ). This implies g(x1 ) = 0, and since x is arbitrary, g ≡ 0. Therefore f ≡ 0 is the general solution in case i). 1 ii) f (0) = . By letting x2 = · · · = xn = 0 in (1) we obtain n f 2 (x1 ) = f 3 (x1 ) +

n−1 n3

and this implies f 2 (x1 + · · · + xn ) = f 2 (x1 ) + · · · + f 2 (xn ) −

depending only on n). Then, by (2)

i.e. k =

(2)

n−1 = y 2 gives y = k (constant, n3

Let f (x1 ) = y. The equation y 3 +

k 2 = nk 2 −

(n − 1)n . n3

(n − 1)n n3

1 1 . Therefore f (x) = in case ii). n n 319

References [1] B. Bartha, OQ.742, Octogon Math. Mag., 9(2001), no.2, 1086.

10

An application of Chebyshev’s inequality We have to determine all functions f : R → (0, ∞) such that  n n X 0 f (xk )   n Y  f 0 (xk )  k=1  ≤ n   X f (xk )   k=1 f (xk ) k=1

for all xk ∈ R (k = 1, n).

f 0 (xi ) = ai (i = 1, n) and f (xi ) suppose that ai > 0. Then by the arithmetic-geometric inequality one can write v u n uY a1 + · · · + an n t ak ≤ . n We will present here a partial solution. Put

k=1

Now, we will study an inequality of type a1 + · · · + an a1 f (x1 ) + · · · + an f (xn ) ≤ n f (x1 ) + · · · + f (xn ) ! ! Ã n Ã n X X 0 f (xk ) / f (xk ) . =

(1)

k=1

k=1

Remark that, when f (x1 ) + · · · + f (xn ) > 0, (1) is equivalent to (a1 + · · · + an )[f (x1 ) + · · · + f (xn )] ≤ n[a1 f (x1 ) + · · · + an f (xn )].

(2)

When the sequences (ai ) and (f (xi )) have the same type of monotony, by Chebysev’s inequality, this is true. Therefore, the following has been proved. Theorem. Suppose that f 0 (xi ) i) > 0, i = 1, n f (xi ) ii) f·(x1 ) + · · · + f (xn¸) > 0 f 0 (xi ) f 0 (xj ) iii) − [f (xi ) − f (xj )] > 0 for all i, j. f (xi ) f (xj ) Then inequality (1) is true. 320

11

On a functional inequality We will describe solutions of the following inequality: µ ¶ x+y 2 x+y f (x)f (y) ≤ f , ∀ x, y, ∈A⊂R 2 2

(1)

(see [1]). Clearly A must be a mid (or Jensen) convex set. Let X = {x ∈ A : f (x) = 0}, Y = {x ∈ A : f (x) > 0}, Z = {x ∈ A : f (x) < 0}. If X 6= ∅, Y = ∅, Z = ∅, then f ≡ 0. If X 6= ∅, Y 6= ∅, Z 6= ∅, then f (x) ≥ 0 and any x ∈ X, y ∈ Y or x, y ∈ X satisfy (1) (so if at least x+y one of x and y is in X). If x, y ∈ Y and Y is mid-convex (i.e. ∈ Y, 2 ¶ µ g(x) + g(y) x+y ≤ , i.e. g too), then g(t) = − ln f (t), t ∈ Y satisfies g 2 2 is Jensen-convex function on Y . For References to these functions, see¶[1]. In µ x+y < 0, case, if Y is not mid-convex, i.e. f (x) > 0, f (y) > 0 but f 2 remark that still (1) can be written as ¯ µ ¶¯ p ¯ x + y ¯¯ ¯ f (x)f (y) ≤ ¯f ¯, 2 so we may take h(t) = − ln |f (t)|, t ∈ Y ∪ Z (since remark that for such x, y one has ¯ µ ¶¯ ¶ µ ¯ 2 x+y ¯ x + y 2 ¯=f |f (x)f (y)| ≤ ¯¯f ) ¯ 2 2 which is J-convex on Y ∪ Z (we have used f (x)f (y) > 0, ∀ x, y ∈ Y ∪ Z). The other possibilities (e.g. X = ∅) are even more simple to handle.

References [1] J. Sándor, On the Open Problem OQ.573, Octogon Math. Mag., 9(2001), 943-944.

12

An unsolved functional equation ”What is the general solution of the functional equation: f {xf (y) + yf (z) + zf (x) − xyz} = f (x)f (y)f (z) + xyz?” Put x = 0. Then f [yf (z) + zf (0)] = f (0)f (y)f (z). 321

Let z = 0 here; implying f [yf (0)] = f 2 (0)f (y). For y = 0 this gives f (0) = f 3 (0), so three possibilities arose: i) f (0) = 0, ii) f (0) = 1, iii) f (0) = −1. In case i) the first equation above yields f (yf (z)) = 0.

(1)

z0 , z = z0 in (1). Then f (z0 ) f (z0 ) = 0, contradiction. Thus f ≡ 0, which is in fact impossible, since then the initial equation would imply 0 = xyz, which is not true for x 6= 0, y 6= 0, z 6= 0. So, in case i) there is no solution. Cases ii) and iii) remain momentary unsolved. Suppose there exists z0 with f (z0 ) 6= 0. Put y =

References [1] Th.M. Rassias, OQ.982, Octogon Math. Mag., 10(2002), no.2, 10411042.

322

Chapter 8

Diophantine equations ”... Wherever there is number, there is beauty.” (Proclus Diadochus)

”... An equation means nothing to me unless it expresses a thought of God.” (Srinivasa Ramanujan)

323

1

Variations on a problem with factorials

1. In the recent issue of Erdélyi Matematikai Lapok [1] F. Smarandache has proposed the following problem: Solve in positive integers the equation: (x!)2 + (y!)2 = (z!)2 .

(1)

Since this is a particular case of the equation X2 + Y 2 = Z2

(2)

where solutions are known to be given by X = d(a2 − b2 ),

Y = 2dab,

Z = d(a2 + b2 )

(3)

where d ≥ 1 is arbitrary, a > b are of opposite parity, and (a, b) = 1; we can use (2) to solve (1). Clearly Z > X and Z > Y , so in (1) z! > y!. This implies z > y, since otherwise z ≤ y would imply z! = 1 · 2 . . . z ≤ 1 · 2 . . . y = y! But then z! = 1 · 2 . . . y(y + 1) · · · = y!K, so y! is a divisor of z!. By (3) we must have that 2dab divides d(a2 + b2 ), i.e. 2ab|(a2 + b2 ), which is impossible, since 2ab is even, while a2 + b2 is odd. Therefore, equation (1) is not solvable in positive integers. 2. There are many possible variations on equation (1). Namely: (x2 )! + (y 2 )! = (z 2 )!

(4)

(x!)2 + (y!)2 = (z 2 )!

(5)

(x2 )! + (y 2 )! = (z!)2

(6)

(x!)2 + (y 2 )! = (z!)2

(7)

(x!)2 + (y 2 )! = (z 2 )!

(8)

In fact, these are the special cases of the following equation: a! + b! = c!

(40 )

a2 + b2 = c!

(50 )

a! + b! = c2

(60 )

324

a2 + b! = c2

(70 )

a2 + b! = c!

(80 )

Remark that in (40 ) a = x2 , b = y 2 , c = z 2 , in (50 ) a = x!, b = y!, c = z 2 , in (60 ) a = x2 , b = y 2 , c = z!, in (70 ) a = x!, b = y 2 , c = z!, in (80 ) a = x!, b = y 2 , c = z 2 . Though, after the study of equations (40 )−(80 ), these particular cases need a special treatment, in what follows, we will restrict our attention to these equations (40 ) − (80 ) only. 3. The equation a! + b! = c! (i.e. (40 )). This is well-known, but we will give for the sake of completeness the solution (see also [3] for the more general equation). Let a, b, c ≥ 1. As in the case of equation (1), here in the same manner, as c > a, c > b, one has a!|c!, b!|c!; let c! = a!M , c! = b!K. Then b! = a!(M − 1), and a! = b!(K − 1), so a!|b! and b!|a!, implying a! = b!. This gives a = b, and 1 = M − 1, 1 = K − 1, i.e. M = 2, K = 2. On base of c! = 2a! c! = (a + 1) . . . c = 2, which is possible only if a = 1, since for a ≥ 2 one has a! clearly (a + 1) . . . c ≥ 3. Thus a = 1, b = 1, c = 2 is the only solution to (40 ). (By the way, since c = 2 = z 2 is not solvable in integers, equation (4) doesn’t have any solution). The equation a2 + b2 = c! (i.e. (50 )). This is a particular case of the equation a2 + b2 = n. It is well-known (see e.g. that [2]) this is possible only if in the prime factorization of n each prime of the form p ≡ 3 (mod 4) appears to an even power (here ”even” includes zero, too). There are many cases when c! satisfies this property, e.g. z! = 24 ·32 ·5·7 (each prime factor is of type p ≡ 1 (mod 4)) 10! = 28 · 34 · 52 · 7. For 12! = 210 · 35 · 52 · 7 · 11 this is not satisfied, since 2 ≡ 3 (mod 4) has an odd power 5. On the other hand, equation (50 ) cannot be solved for infinitely many c. Let c ≡ 3 (mod 4) be a prime. Then clearly in c! this prime appears to an odd power. The equation a! + b! = c2 (i.e. (60 )). First remark that for a = 1 one gets the equation b! + 1 = c2 (9) which is a famous unsolved equation (see e.g. [4]). It is easy to verify that 4! + 1 = 52 , 5! + 1 = 112 , 7! + 1 = 712 . It is conjectured that these are the only solutions to (9) but this seems hopeless at present. Let now 2 ≤ a ≤ b. Though the general case seems difficult, we can completely settle the case b = a + 1. We will show that the only solution of the equation a! + (a + 1)! = c2

(10)

is a = 4, c = 12, i.e. 4! + 5! = 122 . Equation (10) can be written also as 1 · 2 · 3 . . . a(a + 2) = c2 . 325

(∗)

If a + 2 = prime, clearly this is impossible (right side a square, left side a not). Let p the greatest prime between and a + 2 (by Chebyshev’s theorem, 2 a there is such a prime). If a+1 is not a prime, then < p < a, then a < 2p, so p 2 will appear at power 1. This means that (∗) is again impossible. Let therefore a + 1 = q prime, when the equation becomes (q − 1)!(q + 1) = c2 .

(∗∗)

We will show that this is possible only for q = 5. Let p < q − 1 be the greatest prime. If p - (q + 1), clearly (∗∗) will be impossible, as above. Let q − 1 = 2n; then m < p < 2m. By p ≥ m + 1 and q + 1 = 2(m + 1), one can have p = 2 or p = m + 1. Now, for m ≥ 4 it is known that there are at least two prime numbers between m and 2m. Thus p 6= m + 1, by the definition of p. Thus m ≤ 3. The case m = 3 is impossible (p = m + 1 being even), so q−1 = 2, i.e. q = 5. m = 2, when 2 2 Equation a + b! = c2 (i.e. (70 )). This is a particular case of an equation of type c2 − a2 = n (11) (where n = b! in our case). We will show that (11) is solvable if and only if n > 1 is an odd integer, or divisible by 4. Indeed, if n = c2 − a2 , and c, a have the same parity, then m is a multiple of 4. If c and a are of opposite parity, then clearly, n will be odd. Reciprocally, if n = 4m, then n can be written as n = (m + 1)2 − (m − 1)2 . If n = 2m + 1 (odd), then n = (m + 1)2 − m2 . In other words n cannot have the form n ≡ 2 (mod 4). Equation a2 + b! = c! (i.e. (80 )). We will show that this equation can have at most a finite number of solutions. For example, when b = 1, written in the form a2 + 1 = c! (12) has the only solution a = 1, b = 1, c = 2. Indeed, if c ≥ 4, then 4|c!, so a must be odd. Then it is well-known that a2 ≡ 1 (mod 8), so a2 + 1 ≡ 2 (mod 8), which means that 4 - (a2 + 1). Therefore c ≤ 3, and by verifying these cases one easily finds the solutions. Let p be a prime factor of b! such that p2n+1 kb! (m ≥ 0). Such a prime does exist since otherwise each prime would appear to an even power and b! would be a perfect square, which is impossible. Since c > b, then b!|c!, so p2n+1 |c. Clearly one can select c ≥ K (e.g. K ≥ p2m+2 ), such that p2m+2 |c. Now, since p2m+2 |a2 , one has a2 = p2m · p · a0 , so p2m |a2 , giving pm |a, or a = pm · A. This implies p2m A = p2m · p · a0 , thus A2 = pa0 . Since p|A2 , and p is prime, p|A, so A = pA0 , giving a = pm+1 A0 , thus a2 = 326

p2m+2 A0 2 = M p2m+2 . Now a2 + b! = M p2m+2 + p2m+1 B = p2m+1 (M p + B), where p - B, so p2m+1 ka2 + b!. Since p2m+2 |c!, the equation is impossible for c ≥ K. Therefore, we must have a finite number of solutions.

References [1] F. Smarandache, Problem 11136, Erdélyi Matematikai Lapok 5 (2004), no.2, 108. [2] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, Oxford, 1960. [3] J. Sándor, On the Diophantine equation x1 ! + · · · + xn ! = xn+1 !, Octogon Math. Mag., 9(2001), no.2, 963-965. [4] R.K. Guy, Unsolved problems in Number Theory, Springer Verlag, 1994.

2

On the Diophantine equation x1 x2 + x2 x3 + · · · + xn x1 = n + x1 + · · · + xn

This equation has been proposed by Th.M. Rassias in [1]. Remark first that, this equation always has at least a solution in integers, since x1 = 0, x2 = 0, . . . , xn = −n provide a solution. In nonnegative integers a solution is x1 = 0, x2 = 0, . . . , xn−2 = 0, xn−1 = 2, xn = x2 (n ≥ 3); while a solution with all components ≥ 1 is x1 = 1, x2 = 1, . . . , xn−1 = 1, xn = n + 1. Now, we shall prove that equation (1) has at most a finite number of solutions in positive integers; while in the set of all integers, it has infinitely many solutions. By writing (1) for the equation, it can be written as x1 (x2 − 1) + x2 (x3 − 1) + · · · + xn−1 (xn − 1) + xn (x1 − 1) = n

(2)

Clearly for xi ≥ 1, i = 1, n, one must have 0 ≤ x1 (x2 − 1) ≤ n, . . . , 0 ≤ xn (x1 − 1) ≤ n, which imply surely 1 ≤ xi ≤ n + 1, i = 1, n. When, e.g. x1 = 0 and xi ≥ 1, i = 2, n, then by the obvious relation xy ≥ x + y − 1, x, y ≥ 1, one can write successively x2 x3 ≥ x2 + x3 − 1, . . . , xn−1 xn ≥ xn−1 + xn − 1, 327

so by supposing x2 ≤ x3 ≤ · · · ≤ xn−1 ≤ xn , by n + x2 + · · · + xn ≥ x2 + 2x3 + · · · + 2xn−1 + xn − (n − 1) we get x3 + · · · + xn−1 ≤ 2n − 1, i.e. indeed one can have a finite number of x2 , x3 , . . . , xn−1 . For these finite values each time we get an equation axn = b, giving also a finite number of xn ’s. We now show, that in the set of all integers, the number of solutions is infinite. Let, for simplicity n = 3. When x, y, z < 0 in the equation xy + yz + xz = 3 + x + y + z,

(3)

then by putting x = −X, y = −Y , z = −Z one has XY + Y Z + XZ = 3 − (X + Y + Z) < 3

(4)

(where X, Y, Z > 0) having a finite number of solutions. The same happens when x = 0, y < 0, z < 0. Let us now consider the two remaining cases, namely: (i) x < 0, y < 0, z > 0 (ii) x > 0, y > 0, z < 0. Case (ii) is very similar to (i), so we shall study only this latest possibility. Let x = −a, y = −b, z = c. Then a, b, c ≥ 1 and by (3), one gets ab − bc − ac = 3 − a − b + c.

(5)

Let a = λ be arbitrary. Then (5) gives b(λ + 1) + λ − 3 = c(b + λ + 3), i.e. (b + λ + 3)(λ + 1) − [(λ + 1)(λ + 3) − λ + 3] = c(b + λ + 3). Therefore b + λ + 3 must divide (λ + 1)(λ + 3) − λ + 3 = λ2 + 3λ + 6. Since λ2 + 3λ + 6 has always as the divisor λ2 + 3λ + 6, we get for b the value given by b + λ + 3 = λ2 + 3λ + 6, i.e. b = λ2 + 2λ + 3. Thus a = λ, c = λ, b = λ2 + 2λ + 3 give a solution to (5). But all solutions may be determined in this way! Indeed, b + λ + 3 is a divisor of λ2 + 3λ + 6

(6)

so for fixed λ we can obtain a finite number of values of b. The value of c is given by λ2 + 3λ + 6 c=λ+1− , b+λ+3 for b + λ + 3 dividing λ2 + 3λ + 6. Acknowledgement. The author thanks Mihály Bencze for pointing out a fallacy in the first draft of this note. 328


3

On the equation n + (n + 1) + · · · + (n + k) = n(n + k) In what follows we will determine all n, k positive integers such that n + (n + 1) + (n + 2) + · · · + (n + k) = n(n + k)

(1)

(the examples 3 + 4 + 5 + 6 = 3 · 6 and 15 + 16 + 17 + · · · + 35 = 15 · 35 are given in [2]). By simple formulae, (1) can be transformed into 2n2 − 2n − k(k + 1) = 0. Resolving this equation of second degree, one gets p 1 + 1 + 2k(k + 1) n= 2

(2)

(3)

where 1 + 2k(k + 1) = u2 and u is odd. This gives the equation 2k 2 + 2k + 1 − u2 = 0, having the positive solution p −1 + 1 − 2(1 − u2 ) , where 2u2 − 1 = v 2 . k= 2

(4)

(5)

The equation 2u2 − v 2 = 1

(6)

is known as a ”conjugate” Pell equation; namely the conjugate of the classical Pell equation ([3]) u2 − 2v 2 = 1. (7) √ It is well-known that (since 2 is irrational), this equation has infinitely many solutions, and all solutions are determined by the recurrence ½ u1 = u0 um+1 = u0 um + 2v0 vm , vm+1 = v0 um + u0 vm , (8) v1 = v0 where (u0 , v0 ) is the smallest solution of (7). Thus (u0 , v0 ) = (3, 2) give um+1 = 3um + 4vm ,

vm+1 = 2um + 3vm .

329

This gives a sequence of solutions (3, 2); (17, 12); (99, 70), . . . On the other hand, the general solutions of equation (6) are given by u = xm , v = ym , where ½ xm = Aum + Bvm ym = Bum + 2Avm where (un , vm ) are the solutions given in (8), while (A, B) is the smallest solution of (6); i.e. (A, B) = (1, 1) (see e.g. [4], [1]). Thus the solutions of (6) are xm = um + vm , ym = um + 2vm . (9) By (3) and (5), we get n=

1 + xm , 2

k=

ym − 1 , 2

m≥0

(10)

where xm and ym are given by (9). Remark that, by induction it follows from (8) that um is odd and vm is even for all m ≥ 0. By (9) this means that xm is odd and ym is odd for all m ≥ 0, so in (10) n and k are indeed positive integers. Equation (10) gives all possible solutions of (1). For example, x1 = u1 +v1 = 5, y1 = u1 + 2v1 = 7, giving n = 3, k = 3, whence x2 = u2 + v2 = 29, y2 = 41, so n = 15, k = 20. Now x3 = u3 + v3 = 99 + 70 = 169, y3 = 99 + 270 = 239; so n = 85, k = 119. This last example gives the third solution of (1); however all are given by (8), (9) and (10).

References [1] T. Andreescu, D. Andrica, On the resolvation of the equation ax2 −by 2 = 1 in natural numbers (Romanian), Gaz. Mat. 4(1980), 146-148. [2] M. Bencze, OQ.756, Octogon Math. Mag., 9(2001), no.2, 1088. [3] I.M. Vinogradov, Bazele teoriei numerelor, Bucure¸sti, 1954. [4] D.T. Walker, On the Diophantine equation mx2 − ny 2 = ±1, Amer. Math. Monthly, 74(1966), no.5, 504-513.

4

On a problem of Subramanian, and Pell equations

”Find all positive integers x such that both 2x2 + 1 and 3x2 + 1 are simultaneously perfect squares”. Let 2x2 + 1 = X 2 , 3x2 + 1 = Y 2 . This gives 2Y 2 − 3X 2 = −1, so 3X 2 − 2Y 2 = 1 (1) 330

X2 − 1 = 4k, so x is even. 2 Since Y 2 = 3x2 + 1, Y must be odd. Therefore we must find all such solutions of equation (1), where X is odd and Y is odd. Now, the general theory of equations aX 2 − bY 2 = c says (see e.g. [2], [3]) that all its solutions can be written as X = s0 u + bt0 v, Y = t0 u + as0 v, where (u, v) is any solution to X must be odd, so X 2 = 8k + 1, implying x2 =

u2 − abv 2 = c

(∗)

while (s0 , t0 ) is the minimal solution to as2 − bt2 = 1.

(∗∗)

In our case c = 1, a = 3, b = 2, s0 = 1, t0 = 1, so we get X = u + 2v,

Y = u + 3v

(2)

where u2 − 6v 2 = 1.

(∗ ∗ ∗)

Since the minimal solution√to (∗ ∗ ∗) is √ u0 = 5, v0 = 2, the general solution will be obtained from un + √ vn 6 = (5 +√2 6)n (see e.g. [2],√[3]). √ By writing un+1 + vn+1 6 = (5 + 2 6)n+1 = (un + vn 6)(5 + 2 6), we get the recurrence ½ un+1 = 5un + 12vn (3) vn+1 = 2un + 5vn Since u0 = 5, v0 = 2 we get u1 = 49, v1 = 20, etc., and by induction immediately follows un = odd, vn = even for all n ≥ 0. Thus, by (2) we get Xn = un + 2vn = odd, Yn = un + 3vn = odd, so all solutions of the initial problem are found.

References [1] K.B. Subramanian, OQ.1241, Octogon Math. Mag., 11(2003), no.1, 395. [2] D.T. Walker, On the Diophantine equation mx2 − ny 2 = ±1, Amer. Math. Monthly, 74(1966), 504-513. [3] T. Nagell, Introduction to Number Theory, John Wiley and Inc., 1951. 331

5

On the Diophantine equation x3 + y 3 + z 3 = a This note contains certain informations on equations of type: x3 + y 3 + z 3 = a

(1)

particularly when a = 30, we obtain the Open Problem OQ.277 proposed by Th.M. Rassias (Octogon). From the considerations which follow, we can state that when a has the form a = 9m + 4 or a = 9m − 4, the equation (1) have no solutions in integers (m ∈ Z). Thus the equations x3 +y 3 +z 3 = 31 (31 = 9·3 = 4), x3 +y 3 +z 3 = 32 (32 = 9 · 4 − 4), x3 + y 3 + z 3 = 22 (22 = 9 · 2 + 4), etc., none have solutions. When a = 30, unfortunately we can prove only a partial result. 1. For x, y, z < 0, clearly there are no solutions. 2. When x, y, z > 0, there are no solutions, since x, y, z ≤ 3 and x = 0: 13 + 13 = 2, 13 + 23 = 9, 13 + 33 = 28, 23 + 33 > 30; for x = 1 ⇒ y 3 + z 3 6= 29, . . . , x = 3 ⇒ y 3 + z 3 = 3, impossible. 3. Let x < 0, y > 0, z > 0. Put x = −X, when the equation becomes y 3 + z 3 = X 3 + 30,

X, y, z > 0.

(2)

We will prove that this equation has no solutions. Let X = 3k + r, y = 3m + p, z = 3n + t, where r, p, t ∈ {0, 1, 2}. Then we obtain from (2): 27k 3 + 27k 2 r + 9kr2 + r3 + 27m3 + 27m2 p + 9mp2 + p3 = 27n3 + 27n2 t + 9nt2 + t3 + 30.

(3)

This implies: r3 + p3 ≡ t3

(mod 3).

(∗)

i) r = 0. Then from (∗) p = t, and after reducing each term divisible by 3, excepting 10. ii) r = 1. Then 1 + p3 ≡ t3 (mod 3). For p = 0 we have t = 1. Then r = t, and we proceed as above. For p = 1, 2 ≡ t3 (mod 3) is impossible, for p = 3 we can have t = 0, when each term is ≡ (mod 9), excepting 30. iii) r = 2. Then p 6= 0; for p = 1 we have t = 0, for p = 2, t = 1, when 31 − 16 = 15 is not divisible by 9. Thus, in all cases the equation (3) (thus (2)) are not solvable. 332

4. x < 0, y < 0, z > 0. Put x = −X, y = −Y , when we obtain: z 3 = X 3 + Y 3 + 30,

X, Y, z > 0.

(4)

Let z = 3k + r, X = 3m + p, Y = 3n + t, r, p, t ∈ {0, 1, 2}. Then we get from (4): 27k 3 + 27k 2 r + 9kr2 + r3 = 27m3 + 27m2 p + 9mp2 + p3 + 27n3 + 27n2 t + 9nt2 + t3 .

(5)

This implies: r3 ≡ p3 + t3

(mod 3).

(∗∗)

By analyzing all cases as above, in the case r = 1, p = 2 we obtain t = 2. This case is possible, since in (5) we deduce 30 + 16 − 1 = 45, which is ≡ 0 (mod 9). By dividing with 9 each term in (5), we can deduce: 3k 3 + 3k 2 + k = 3m3 + 6m2 + 4m + 3n3 + 6n2 + 4n + 5.

(6)

Thus, we have obtained that when (4) is solvable, then necessarily z = 3k +1, X = 3m+2, Y = 3n+2, with k, m, n satisfying (6). We have considered many particular cases, and many computations, but were unable to prove that (6) has no solutions. The author feels that the equation has no solutions at all. Thus we make the conjecture that the equation x3 + y 3 + z 3 = 30 has no solutions in integers. (This is equivalent that (6) is not solvable in positive integers k, m, n).


6

On the Diophantine equation x3 + y 3 + z 3 = a, II In the first part [12], we have considered equations of type x3 + y 3 + z 3 = a

(1)

and a special study was denoted to the particular case a = 30. This was proposed as an OQ.277 [13]. We have reduced the study of equation to x < 0, y < 0, z < 0. We note have that this Open Problem (when a = 30) is included also in W. Sierpinski’s famous book [11] or L.J. Mordell’s monograph [10]. In [14] Z. Tuzson published an erroneous proof that this equation has no solution (the fallacy in the proof follows at one by remarking that on 333

p.219 of [14] in relation (1) we must have 30 − 3xyz (and not 30 + 3xyz)) and as a result, 10 − S3 in relation (2) (not 10 + S3 )! This invalidates the argument what follows. Equations of type (1) (and particularly, with a = 30) have a long history. Practically, today theoretical methods are combined with computer search algorithms. In finding all solutions for a range of values of a with max{|x|, |y|, |z|} ≤ A, a straightforward two-dimensional algorithm [3], [6], [9] was applied in O(A2 ) steps. In [6], a computer search based on this algorithm for max{|x|, |y|, |z|} ≤ 221 − 1, 1 ≤ n ≤ 999 was discussed. All 5418 solutions were deposited into the UMT file of American Math. Society. Particularly, the search gives solutions for 17 values of a for which no solutions has been found before: a ∈ {39, 143, 180, 231, 312, 321, 367, 438, 462, 516, 542, 556, 660, 663, 754, 777, 870}. Recently (see [7]) for a = 439 the solution (−869418, −2281057, 2322404) and for a = 462 the solution (1612555, 2598019, −2790488) were obtained. For a = 478 (see [7]) the solution (−1368722, −13434503, 13439237) √ was found. In [5] a new algorithm based on the class number of Q( 3 a) was applied. On a CYBER 205 vector computer for many valued of a in the range max{|x|, |y|, |z|} ≤ A no solutions has been found. Particular values of a include: a = 30, 33, 42, 52, 74, 75, 110, 114, 156, 165, 195, 290, 318, . . . , so for all these equations we could conjecture that they have no solutions at all. In a recent paper new solutions for a = 75, 435, 444, 501, 600, 618, 912, 969 in the range |x| = min{|x|, |y|, |z|} ≤ 2 · 107 were found. For history or other aspects of these equations, see other titles in the References. Equations of such type appear also in [15]. 334

References [1] A. Bremner, On sums of three cubs, Canadian Math. Soc. Conf. Proc., 15(1995), 87-91. [2] B. Conn, L. Vaserstein, On sums of three integral cubs, Contemp. Math., 166(1994), 285-294. [3] V.L. Gardiner, R.B. Lazarus, P.R. Stein, Solutions of the Diophantine equation x3 + y 3 = z 3 − d, Math. Comp., 18(1964), 408-413. [4] R.K. Guy, Unsolved problems in Number Theory, Springer, New York, 1994 (second edition). [5] D.R. Heath-Brown, W.M. Lioen, H.J.J. teRiele, On solving the Diophantine equation x3 + y 3 + z 3 = k on a vector processor, Math. Comp., 61(1993), 235-244. [6] K. Koyama, Tables of solutions of the Diophantine equation x3 +y 3 +z 3 = n, Math. Comp., 62(1994), 941-942. [7] K. Koyama, Y. Tsuruoka, H. Sekigawa, On searching for solutions of the Diophantine equation x3 +y 3 +z 3 = n, Math. Comp., 66(1997), 841-851. [8] R.F. Lukes, A very fast electronic number sieve, Ph.D. Thesis, Univ. of Manitoba, 1995. [9] J.C.P. Miller, M.F.C. Woollett, Solutions of the Diophantine equation x3 + y 3 + z 3 = k, J. London Math. Soc., 30(1955), 101-110. [10] L.J. Mordell, Diophantine equations, Academic Press, 1969. [11] W. Sierpinski, A selection of problems in the theory of numbers, Pergamon Press, 1964. [12] J. Sándor, On a Diophantine equation, Octogon Math. Mag., 8(2000), no.1, 221-222. [13] Th.M. Rassias, OQ.277, Octogon Math. Mag., 7(1999), no.2. [14] Z. Tuzson, A solution to OQ.277, Octogon Math. Mag., 8(2000), no.1, 219-220. [15] M. Bencze, OQ.340, Octogon Math. Mag., 8(2000), no.1, 273. 335

7

n dimensional cuboids with integer sides and diagonals

In what follows we will determine all n dimensional cuboids with integer sides a1 , a2 , . . . , an , having integer diagonals. Clearly, this leads to the Diophantine equation a21 + a22 + · · · + a2n = a2 .

(1)

Let d = (a1 , a2 , . . . , an ), i.e. ai = dxi (i = 1, n), where (x1 , xn ) = 1. Then from (1) we get that d2 |a2 . It is well-known, that then this implies d|a. Let a = dx. Then (1) becomes: x21 + x22 + · · · + x2n = x2

(2)

where (x1 , x2 , . . . , xn ) = 1. This implies (x1 , x2 , . . . , xn , x) = 1, too. Let x1 = ky1 , . . . , xn−1 = kyn−1 , xn = kyn − x, for integers y1 , . . . , yn , and certain rational k. Then (2) becomes: k(y12 + y22 + · · · + yn2 ) = 2xyn .

(3)

This gives 2xn yn = 2(kyn − x)yn = 2kyn2 − 2xyn 2 = 2kyn2 − k(y12 + · · · + yn2 ) = k(yn2 − y12 − · · · − yn−1 ).

Thus: 2 k(yn2 − y12 − · · · − yn−1 ) = 2xn yn .

(4)

(3) and (4) imply x 1 = y1

yn2

2xn yn 2xn yn , . . . , xn−1 = yn−1 2 2 2 2 2 yn − y1 − · · · − yn−1 − y1 − . . . yn−1

and xn = kyn − i.e.

2 ) k(yn2 − y12 − · · · − yn−1 k(y12 + · · · + yn2 ) = , 2yn 2yn

x1 xn−1 xn = ··· = = 2 2 2 2y1 yn 2yn−1 yn yn − y1 − · · · − yn−1 =

y12

x k . = 2 2yn + · · · + yn 336

(5)

Let us suppose that

p k is the reduced form of (i.e. (p, q) = 1). Then q 2yn

q q q 2 2y1 yn = x1 , . . . , 2yn−1 yn = xn−1 , yn2 − y12 − · · · − yn−1 = xn , p p p q y12 + · · · + yn2 = x. p Since on the left sides we have integers and (q, p) = 1, thus p|x, p|xn , . . . , p|xn−1 , p|x1 , which is possible only when p = 1, since (x, xn , . . . , xn−1 , x1 ) = 1. Thus we have obtained that if (x1 , . . . , xn , x) is a solution of equation (2), then there exist integers y1 , . . . , yn and q > 0 such that qx1 = 2y1 yn , qx2 = 2y2 yn , . . . , qxn−1 = 2yn−1 yn , qxn = yn2 − y12 − · · · − yn2 , (6) qx = y12 + · · · + yn2 . Here q > 0 is taken so that (x1 , . . . , xn ) = 1. Conversely it is easy to see that if x1 , . . . , xn , x satisfy the equations (6) with positive integers y1 , . . . , yn and q > 0, integer, then we have obtained a solution for (2), if (y1 , . . . , yn ) = 1. Thus, the general solution of equation (1) is given by a1 = dx1 = d

2y2 yn 2yn−1 yn 2y1 yn , a2 = dx2 = d , . . . , an−1 = d , q q q

an = d

2 yn2 − y12 − · · · − yn−1 y 2 + · · · + yn2 , a=d 1 q q

(7)

where yi (i = 1, n) and q are as above.

8

The equation {x2 } + {y 2 } = {z 2 }

Let {x} denote the fractional part of the real number x. We will solve the equation {x2 } + {y 2 } = {z 2 }. (1) Since x2 = a has solutions only when a ≥ 0, and x = ±a, clearly it will be sufficient to solve the equation {a} + {b} = {c}

(2)

in nonnegative real numbers. Then if (a, b, c) is a solution of (2), this will generate eight solutions of (1) (i.e. (a, b, c); (−a, b, c); (a, −b, c); (a, b, −c); 337

(−a, −b, c); (a, −b, −c); (−a, b, −c); (−a, −b, −c)). We prove that (2) has solutions only if {a} + {b} < 1. Indeed, if (2) is true, then {c} = c − [c] < 1, so {a} + {b} < 1. Reciprocally, if θ = {a} + {b} < 1 (clearly θ ≥ 0), then the equation θ = {c} is solvable in c (in fact in each interval [m, m + 1) (m ≥ 0, integer), there is a single solution c). Therefore, we must study the inequality {a} + {b} < 1.

(∗)

6

y = {x}

1 y=0 O

-

1

2

3

n n+1

Let a ∈ [m, m + 1), b ∈ [n, n + 1). If a ∈ [m, m + 1/2), b ∈ [n, n + 1/2), then 1 1 {a} + {b} = (a − m) + (b − n) < + = 1, so (∗) is true. But this condition 2 2 is only sufficient. One may happen that (∗) is true for a ∈ [m, m + 1/2), b ∈ [n + 1/2, n + 1) or a ∈ [m + 1/2, m + 1), b ∈ [n, n + 1/2) (e.g. m = n = 0, 3 1 a = , b = . Then a < 1/2, b > 1/2 and a + b < 1). In fact, (∗) can be 5 5 written equivalently as a + b < m + n + 1. (3) Equation (2) may be generalized for three or more variables, but their study follows the same line as above.

9

On the Diophantine equation

1 1 1 a + +· · ·+ = x1 x2 xn b

1. We will study the equation in the title, where the unknowns xi (i = 1, n) are positive integers, while a, b are given positive integer numbers. For a = 6, 338

b = n2 −1, this contains the OQ.1119 [1] (by putting x1 = y1 , x2 = y1 +y2 , . . . , xn = y1 + y2 + · · · + yn ). The particular case n = 3 of this Open Problem will b be studied in detail. Clearly, for n = 1 one has x1 = , which is integer only a if a|b. Let n = 2. The equation 1 a 1 + = x1 x2 b

(1)

has been studied in detail in [2]. See also [3]. We will show that the resolvation of the general equation reduces inductively to the resolvation of equations of type (1). Let n = 3. We may suppose that x1 ≤ x2 ≤ x3 . Thus, if 1 1 a 1 + + = , x1 x2 x3 b

(2)

a 3 3b ≤ , implying x1 ≤ . This means that x1 can take a finite number b x1 a · ¸ 3b values). Let x0 be such a value. Then of values (at most a

then

1 a 1 ax0 − b A 1 + = − = = . x2 x3 b x0 bx0 B Thus

1 1 A + = x2 x3 B

(3)

which is an equation of type (1). Thus, in order to solve equation (2) we must solve a finite number of equations of type (3). For general n we can assume n nb a , which gives x1 ≤ , i.e. a finite number of x1 ≤ x2 ≤ · · · ≤ xn . Thus ≤ b x1 a values. For each fixed value of x1 one obtains an equation of n − 1 unknown. 1 1 A + ··· + = , x2 xn B where x2 ≤

(4)

(n − 1)B , etc.; inductively all will be reduced to equations of type A

(1). 2. We will take as an example the case n = 3 of OQ.1119, i.e. the equation 1 1 3 1 + + = . x x+y x+y+z 4 339

(5)

Clearly x ≥ 2; while for x ≥ 4 one has x + y ≤ 5, x + y + z ≥ 6, so the 1 1 1 37 3 37 . Now, ≤ , i.e. 180 ≤ 148 is sum of the left side is ≤ + + = 4 5 6 60 4 60 impossible. Therefore, we must have x > 1, x < 4, i.e. x ∈ {2, 3}. 3 1 1 a) Let x = 2. By − = one obtains the equation 4 2 4 1 1 1 + = (6) t t+z 4 where y + 2 = t. By elementary transformations from (6), we get the equation t2 + t(z − 8) − 4z = 0. (7) √ 8 − z ± z 2 + 64 This gives t1,2 = , and since t > 0, 2 √ 8 − z + z 2 + 64 t= . (8) 2 Here z 2 + 64 = u2 , n ∈ N, otherwise t is irrational. Writing 64 = (u − z)(u + z), and considering the divisors 1, 2, 4, 8, 16, 32, 64 of 64, by an easy verification follows u = 10, z = 6; respectively u = 17, z = 15. By (8) these give t = 6, respectively t = 5. Thus all solutions of equation (6) are: z = 6, y = 4 and z = 15, y = 3. (9) 3 1 5 b) x = 3. Now, by − = we get the equation 4 3 12 1 1 5 + = , (10) t t+z 13 where t = y + 3. We could use the method of a) however we will follow another way, i.e. by solving an equation of type (see [2]) 1 1 5 + = . (11) a b 12 Since 12(a + b) = 5ab, put (a, b) = d, i.e. a = dA, b = dB, with (A, B) = 1. Then 12(A + B) = 5dAB, and by the known fact that (AB, A + B) = 1, it follows that AB|12, i.e. AB ∈ {1, 2, 3, 4, 6, 12}. But (5, 12) = 1 implies 5|(A + B), so we immediately get the solutions A = 1, B = 4; A = 2, B = 3. Then d = 3 or d = 2 and we obtain a = 3 · 1 = 3, b = 3 · 4 = 12; respectively a = 2 · 2 = 4, b = 3 · 2 = 6. From this we can deduce the single solution of (10), namely y = 1, z = 2. (12) By concluding, all solutions of the equation (5) are the following: x = 2, y = 4, z = 6;

x = 2, y = 3, z = 15; 340

x = 3, y = 1, z = 2.

References [1] M. Bencze, OQ.1119, Octogon Math. Mag., 10(2002), no.2, 1075. [2] J. Sándor, On two Diophantine equations (Hungarian), Mat. Lapok, Cluj, 8/2001, 285-286. [3] J. Sándor, Geometric theorems, Diophantine equations and arithmetic functions, American Research Press, New Mexico, 2002.

10

Harmonic triangles

1. Let ABC be a triangle with side lengths a, b, c. If C = 60◦ , problem US4 presented to the XXth IMO asked for a proof of relation c c + ≥ 2. a b

(1)

For (1) quite complicated proofs were given. We note here that, by using elementary means, in fact an easier proof of a stronger relation can be deduced. Indeed, (1) can be written also as H(a, b) ≤ c,

(2)

where H(a, b) denotes the harmonic mean of a and b. Remark that, the stronger inequality a+b ≤c (3) A(a, b) = 2 is valid. Indeed, a + b ≤ 2c ⇔ (a + b)2 ⇔ 4c2 , or a2 + 2ab + b2 ≤ 4c2 = 4(a2 +b2 −ab), since c2 = a2 +b2 −2ab cos 60◦ = a2 +b2 −ab. The last inequality becomes 3(a − b)2 ≥ 0, which is obvious. Therefore H(a, b) ≤ G(a, b) ≤ A(a, b) ≤ c

(4)

is valid in any triangle ABC having C = 60◦ . Since a = 2R sin A, etc., and sin A + sin B = 2 sin

A−B A+B cos , 2 2

sin C = 2 sin

C C cos , 2 2

we can remark that, in any triangle ABC, relation (3) is valid iff cos

A−B C ≤ 2 sin . 2 2 341

(5)

C 1 The case sin = is a particular case of (5). 2 2 2. Now, let us suppose that ABC has integral sides. We say that the triangle ABC is harmonic relatively to the vertex C, if H(a, b) = c.

(6)

We say that the triangle ABC is harmonic, if H(a, b, c) = integer, (7) ¶ µ 1 1 1 denotes the harmonic mean of a, b, c. In where H(a, b, c) = 3/ + + a b c what follows we shall study these two types of harmonic triangles. Relation (6) can be written also as 2ab = (a + b)c

(8)

Let gcd(a, b) = d, i.e. a = da1 , b = db1 , with (a1 , b1 ) = 1. Then (8) becomes 2da1 b1 = (a1 +b1 )c. This implies a1 b1 |(a1 +b1 )c, and since it is well-known that for (a1 , b1 ) = 1 one has also (a1 + b1 , a1 b1 ) = 1, we get a1 b1 |c, so c = ka1 b1 . k(a1 + b1 ) This in turn implies 2d = k(a1 + b1 ), so d = , where at least one of 2 k and a1 + b1 is an even number. From this we get a=

ka1 (a1 + b1 ) , 2

b=

kb1 (a1 + b1 ) , 2

c = ka1 b1

(9)

with (a1 , b1 ) = 1 (and at least one of k and a1 + b1 = even). Now a + b > c, i.e. k(a1 + b1 ) (a1 + b1 ) > ka1 b1 ⇔ (a1 + b1 )2 > 2a1 b1 is always true. The other 2 inequalities b + c > a and a + c > b lead to relations of type a21 + 2a1 b1 > b21 and b21 + 2a1 b1 > a21 so −2a1 b1 < a21 − b21 < 2a1 b1 , i.e. |a21 − b21 | < 2a1 b1 . All in all, the triangle ABC is harmonic relatively to C, if its sides are given by (9), where (a1 , b1 ) = 1; k or a1 + b1 are even; and |a21 − b21 | < 2a1 b1

(∗)

For example, when a1 = b1 + 1 and k = even, the inequality (∗) is true, since 2b1 + 1 < 2b1 (b1 + 1) ⇔ 1 < 2b21 ≥ 2. 3. Now, we shall determine all harmonic triangles of type 2, i.e. for which (7) is true, i.e. X ab|3abc. (10) 342

Let d = gcd(a, b, c), i.e. a = da1 , b³= db1 , c = dc1 ´where (a1 , b1 , c1 ) = 1. X X Then a1 b1 |3da1 b1 c1 . Let D = gcd a1 b1 , a1 b1 c1 . Then a1 b1 c1 = Dk, X a1 b1 = Dk 0 , where (k, k 0 ) = 1 so Dk 0 |3dDk, i.e. k 0 |3dk. This implies k 0 |3d, X X X m a1 b1 so a1 b1 = Dk 0 |3dD. Let 3dD = m a1 b1 , i.e. d = . This gives 3D X X m a1 b1 m a1 b1 a = da1 = a1 , b = db1 = b1 , 3D 3D X m a1 b1 . c = cc1 = c1 3D X Here D| a1 b1 , so for a, b, c to be integers, we have two possibilities: 1) 3|m. Let m = 3s. Then a=

sa1

X D

a1 b1

,

b=

sb1

X D

a1 b1

,

c=

sc1

X

a1 b1

D

.

Clearly a + b > c if a1 + b1 > c1 . Therefore (a1 , b1 , c1 ) = 1 must satisfy also the inequalities a1 + b1 > c1 , b1 + c1 > a1 , a1 + cX 1 > b1 . 2) 3 - m. Remark that we cannot have 3 a1 b1 , since then we must X have 3|a1 , 3|b1 , 3|c1 , contradiction to (a1 , b1 , c1 ) = 1. Thus we have 3| a1 b1 (and again, of course a1 + b1 > c1 , etc.). These two cases determine all harmonic triangles.

11

A Diophantine equation involving Euler’s totient The equation in the title is ϕ(x1 x2 . . . xn ) = ϕ(x1 ) + ϕ(x2 ) + · · · + ϕ(xn ).

(1)

The aim of this Note is to prove that equation (1) has a finite number of solutions. For particular n, all solutions can be obtained by verifications. The method is based on two simple lemmas. Lemma 1. For all positive integers x and y one has ϕ(xy) ≥ ϕ(x)ϕ(y) 343

(2)

with equality only when x and y are coprime. Proof. This inequality is well known, see e.g. [3]. In fact, if one denotes by P = product of all common prime divisors of x and y, then immediately follows the identity P ϕ(x)ϕ(y), ϕ(xy) = ϕ(P ) where P ≥ ϕ(P ). This may be proved e.g. by ¶ Y µ 1 ϕ(x) = x 1− . q q/x,prime

Lemma 2. For n ≥ 2 and ui > 2 positive integers (i = 1, 2, . . . , n), one has u1 u2 . . . un > u1 + u2 + · · · + un .

(3)

Proof. For n = 2 this follows from (u1 −1)(u2 −1) > 1 (i.e. u1 u2 > u1 +u2 ), valid for u1 > 2, u2 > 2. Now, by admitting that (3) is true for n, by (u1 . . . un )un+1 > (u1 + · · · + un )un+1 > u1 + · · · + un + un+1 since this is equivalent to (u1 + · · · + un − 1)(un+1 − 1) > 1. So, by induction, relation (3) is valid for all n ≥ 2. Now, first consider the case n = 2. By (2) and the notations ϕ(x1 ) = u, ϕ(x2 ) = v one can write: u + v ≥ uv, i.e. (u − 1)(v − 1) ≤ 1. We have two cases a) (u − 1)(v − 1) = 0 b) (u − 1)(v − 1) = 1. In case a) we have u = 1 or v = 1. Thus ϕ(x1 ) = 1, i.e. x1 = 1 or x1 = 2, in which case ϕ(x2 ) = 1 + ϕ(x2 ) (which is impossible), or ϕ(2x2 ) = 1 + ϕ(x2 ), for x2 = odd this cannot be true, since then ϕ(2x2 ) = ϕ(2)ϕ(x2 ) = ϕ(x2 ) 6= 1 + ϕ(x2 ). If x2 is even, let x2 = 2k m, with m = odd. Then ϕ(2x2 ) = ϕ(2k+1 m) = k+1 ϕ(2 )ϕ(m) = 2k ϕ(m) = 1 + ϕ(2k m), only if 2k ϕ(m) = 1 + 2k−1 ϕ(m), or 2k−1 ϕ(m) = 1. Then k = 1 and ϕ(m) = 1. Thus m = 1 or 2, and clearly we have determined all solutions. In case b) we have u = 2 and v = 2. Thus ϕ(x1 ) = 2 and ϕ(x2 ) = 2. Then ϕ(x1 x2 ) = 4 etc. 344

Now, let n = 3, i.e. consider the equation ϕ(x1 x2 x3 ) = ϕ(x1 ) + ϕ(x2 ) + ϕ(x3 ). By (2) one has clearly ϕ(x1 x2 x3 ) ≥ ϕ(x1 )ϕ(x2 )ϕ(x3 ), and by (3) we cannot have ϕ(xi ) > 2 for all i = 1, 2, 3. Let ϕ(xi ) = ui . Then, let u1 ∈ {1, 2}. For u1 = 1 one gets 1 + u2 + u3 ≥ u2 u3 , i.e. (u2 − 1)(u3 − 1) ≤ 2. Thus one have to distinguish the cases: a) (u2 − 1)(u3 − 1) = 1 b) (u2 − 1)(u3 − 1) = 2 and in each case one arrives to a finite number of solutions, and these solutions may be obtained by direct verifications. For u1 = 2, one gets 2 + u2 + u3 ≥ 2u2 u3 , implying u2 + u3 ≥ 2(u2 u3 − 1) > u2 u3 if u2 u3 > 2. Thus (u2 − 1)(u3 − 1) < 1. Here it is easy to study all possibilities to deduce the finite number of possible solutions. In the general case one can follow the same argument. Namely, let ϕ(xi ) = ui . By the inequality ϕ(x1 . . . x2 ) ≥ ϕ(x1 ) . . . ϕ(xn ) (which is a consequence of Lemma 1 with induction), and by relation (3) one can deduce that there exists i0 ∈ {1, . . . , n}, with xi0 ∈ {1, 2}. Let i0 = 1 (for simplification of notation), when ϕ(x1 ) = 1, then x1 ∈ {1, 2} and one obtains ϕ(x2 . . . xn ) = 1+ϕ(x2 )+· · ·+ϕ(xn ) or ϕ(2x2 . . . xn ) = 1+ϕ(x2 )+· · ·+ϕ(xn ). From 1+ϕ(x2 )+ · · · + ϕ(xn ) ≥ ϕ(x2 ) . . . ϕ(xn ) we have that there exists j0 ∈ {2, . . . , n} with ϕ(xj0 ) ∈ {1, 2, 3}. (Since we can prove, completely analogously with Lemma 2, that for ui > 3, 1 + u2 + · · · + un < u2 . . . un for n ≥ 2). When ϕ(x1 ) = 2, we have 2+ϕ(x2 )+· · ·+ϕ(xn ) ≥ 2ϕ(x2 ) . . . ϕ(xn ), implying ϕ(x2 )+· · ·+ϕ(xn ) ≥ 2[ϕ(x2 ) + · · · + ϕ(xn ) − 1] > ϕ(x2 ) . . . ϕ(xn ), if ϕ(x2 ) . . . ϕ(xn ) > 2. By Lemma 2, this is impossible if all ϕ(xi ) > 2 for i ∈ {2, . . . , n}. Thus there exists i1 ∈ {2, . . . , n} with ϕ(xi1 ) ∈ {1, 2}. In all cases, we obtain an equation with n − 1 arguments, then two (or three) equations with n − 2 arguments, and so on. Since the equation ϕ(x) = k can have only a finite number of solutions √ (since, e.g. for x > 6, by the known inequality ϕ(x) > x it follows x < k 2 ), finally all equations can have a finite number of solutions. Indeed, one can prove: Theorem. The equation (1) has at most a finite number of solutions. Proof. By (2), equation (1) implies u1 + u2 + · · · + un ≥ u1 u2 . . . un

(4)

where ui ≥ 1 (i = 1, 2, . . . , n) are positive integers, ui = ϕ(xi ). First we prove that inequality (4) has a finite number of solutions (u1 , . . . , un ). Let u1 ≤ u2 ≤ · · · ≤ un . Then (4) yields u1 u2 . . . un ≤ nun or u1 u2 . . . un−1 ≤ n. Thus 1 ≤ u1 ≤ u2 ≤ · · · ≤ un−1 ≤ n, which means that the numbers u1 , . . . , un−1 can take at most the values 1, 2, . . . , n. For un we 345

u1 + · · · + un−1 , u1 u2 . . . un−1 − 1 if u1 u2 . . . un−1 > 1 (if not, then u1 = · · · = un−1 = 1, and we obtain the equation ϕ(x1 . . . xn ) = n − 1 + ϕ(xn ) which we study separately. Thus un can take a finite number of values, too. Since the equation ϕ(x) = k (k fixed) can have at most a finite number of solutions, the proof (in this case) is completed. For the equation ϕ(x1 . . . xn ) = n−1+ϕ(xn ) we note that by ϕ(x1 . . . xn ) ≥ ϕ(x1 . . . xn−1 )ϕ(xn ) we can deduce ϕ(xn )[ϕ(x1 . . . xn−1 ) − 1] ≤ n − 1 thus, if ϕ(x1 . . . xn−1 ) 6= 1, ϕ(xn ) ≤ n − 1. If ϕ(x1 . . . xn−1 ) = 1, then x1 . . . xn−1 ∈ {1, 2}, and the proof is finished. For Diophantine equations involving various arithmetical functions we quote [1], [2]. have un (u1 u2 . . . un−1 − 1) ≤ u1 + u2 + · · · + un−1 , i.e. un ≤

References [1] C.A. Nicol, Some Diophantine equations involving arithmetic functions, J. Math. Anal. Appl., 15(1966), 154-161. [2] J. Sándor, Some Diophantine equations for particular arithmetic functions, Seminarul de teoria structurilor, no.53, Univ. Timi¸soara, Romania, 1989, 1-10. [3] J. Sándor, Some arithmetic inequalities, Bulletin Number Theory Rel. Topics, 11(1987), 149-161.

12

On f (n) + f (n + 1) + · · · + f (n + k) = f (n)f (n + k) for f ∈ {ϕ, ψ, σ}

Our aim is to study the equations in the title in positive integers n, and particular values of k for the arithmetical functions ϕ (Euler’s totient), ψ (Dedekind’s function) and σ (sum of divisors). Theorem 1. The only solution in positive integers of the equation ϕ(n) + ϕ(n + 1) + ϕ(n + 2) = ϕ(n)ϕ(n + 2)

(1)

is n = 3. Proof. An easy computation shows that for n < 19, the only solution of (1) is n = 3: ϕ(3) + ϕ(4) + ϕ(5) = ϕ(3)ϕ(5), i.e. 2 + 2 + 4 = 2 · 4. Let now n ≥ 19. We note that for such n one has ϕ(n) >

√ n + 2.

346

(2)

This can be proved by many arguments. For example, in [2] it is proved that ϕ(n) > 6n/(12 + 5 log n). √ Now, the inequality 6n/(12 + 5 log n) > n + 2 becomes ¶ µ √ 24 10 log n 2 (12 + 5 log n) = 12 + √ + 5 log n + √ 6 n> 1+ √ n n n √ and this is true for n ≥ 19. Here one has n > log n for n ≥ 4, but we need a √ 24 24 slightly stronger 6 n > 15 log n (n ≥ 11), and since 12 + √ < 12 + = 20, 3 n the above inequality follows. This proves (2). Now, (1) can be written as [ϕ(n) − 1][ϕ(n + 2) − 1] = ϕ(n + 1) + 1.

(3)

On the right side of (3) one has ϕ(n + 1) + 1 ≤ n + 1, but on the left side, for n ≥ 19 one can write that √ √ [ϕ(n) − 1][ϕ(n + 2) − 1] > ( n + 1)( n + 2 + 1) p √ √ = n(n + 2) + n + n + 2 + 1 > n + 1, i.e. n2 + 2n + 1 + a > n2 + 2n + 1 for a > 0. This finishes the proof of Theorem 1. Theorem 2. None of the equations ψ(n) + ψ(n + 1) + ψ(n + 2) = ψ(n)ψ(n + 2)

(2)

σ(n) + σ(n + 1) + σ(n + 2) = σ(n)σ(n + 2)

(3)

and has solutions in positive integers. Proof. Write (2) in the form [ψ(n) − 1][ψ(n + 2) − 1] = ψ(n + 1) + 1.

(4)

k(k + 1) 2 (see e.g. [3]) on the left side of (4) we have ≥ n(n + 2), while on the right side of (4) one has (n + 1)(n + 2) ψ(n + 1) + 1 ≤ + 1. 2 Since ψ(n) ≥ n + 1, n ≥ 2, and ψ(k) ≤ σ(k) ≤ 1 + 2 + · · · + k =

347

Therefore, if (4) is true, then n(n + 2) − 1 ≤

(n + 1)(n + 2) , 2

(5)

i.e. n2 + n ≤ 4. Since for n ≥ 2, n(n + 1) ≥ 6, this is impossible. For n = 1, (4) doesn’t give a solution. The proof of (3) runs on the same lines. Remark 1. The proof shows that for any arithmetical function f such that n(n + 1) f (1) = 1, f (n) ≥ n + 1, f (n) ≤ , n≥2 2 the equation f (n) + f (n + 1) + f (n + 2) = f (n)f (n + 2) doesn’t have a solution in positive integers. Remark 2. Equations (1) and (2) are particular cases of OQ.753 and OQ.754 (t = 1, k = 2) see [1]. Theorem 3. The equations ψ(n) + ψ(n + 1) + ψ(n + 2) + ψ(n + 3) = ψ(n)ψ(n + 3)

(6)

σ(n) + σ(n + 1) + σ(n + 2) + σ(n + 3) = σ(n)σ(n + 3)

(7)

and do not have solutions in positive integers. Proof. Now the method of proof of Theorem 2 doesn’t work. We need the following auxiliary results: ψ(n) ≤ σ(n) < 2n log n for n ≥ 3.

(8)

The result is well-known, but we give here the simple proof: σ(n) = n

X1 d|n

d

≤n

X1 d≤n

s

< 2n log n,

since it is well-known that 1+

1 1 + · · · + < 2 log n for n ≥ 3. 2 n

Now, (6) can be written equivalently as: [ψ(n) − 1][ψ(n + 3) − 1] = ψ(n + 1) + ψ(n + 2) + 1. 348

(8)

The left side of (8) is ≥ n(n + 3) for n ≥ 2, but the right side is < 2(n + 1) log(n + 1) + 2(n + 2) log(n + 2) + 1, so we must have µ ¶ µ ¶ 1 2 1 n+3 f (n) + f (n + k). Indeed, (f (n) − 1)(f (n + k) − 1) ≥ 0 ⇒ f (n)f (n + k) ≥ f (n) + f (n + k) − 1 ⇒ 2f (n)f (n + k) ≥ 2f (n) + 2f (n + k) − 2 > f (n) + f (n + k) since f (n) + f (n + k) ≥ 3, by assumption. It is easy to see that for n ≥ 3 one has ϕ(n) ≥ 2, ϕ(n + 1) ≥ 2, ψ(n) ≥ 2, ψ(n + 1) ≥ 2, S(n) ≥ 2, S(n + 1) ≥ 2, so only the cases n ∈ {1, 2} should be considered. Since (1) implies f (n + k)|f (n) (for t = 1) and ϕ(1) = ϕ(2) = 1; ψ(1) = 1, ψ(2) = 3, S(1) = 1, S(2) = 2 we should have ϕ(1 + k) = 1, ϕ(2 + k) = 1, ψ(1 + k) = 1, ψ(2 + k) = 1, ψ(2 + k) = 3, S(1 + k) = 1, S(2 + k) = 2. Only for ϕ the k = 1 case is acceptable, but then ϕ(1)ϕ(2) = 1 6= ϕ(1) + ϕ(2). Therefore, for t = 1 the proposed equations does not have solutions.

14

Some equations involving the arithmetical functions ϕ, σ, ψ WeX will consider the following equations: X n 1) ϕ(d ) = ϕ(nd ) d|n

d|n

350

2)

X

(σ(d))d =

d|n,d6=n

3)

X

X

(σ(d))n−d

d|n,d6=n X ψ(d) = ψ(n − d).

d|n,d6=n

d|n,d6=n

In what follows all equations (1)-(3) will be solved. 1) Let d1 = 1 < d2 < · · · < dr = n be all divisors of n. Then ϕ(1n ) + ϕ(dn2 ) + · · · + ϕ(dnr ) = ϕ(nd1 ) + · · · + ϕ(ndr )

(1)

is impossible for n ≥ 3, since it is well known that for x ≥ 3, ϕ(x) is an even number. Now di ≥ 1, i = 1, r, and on the right side of (1) one has an even number. On the other hand, the left side of (1) is odd, since dn2 ≥ 2n ≥ 8, so ϕ(dn2 ), . . . , ϕ(dnr ) all are even, while ϕ(1n ) = 1 is odd. For n = 1 or n = 2, by ϕ(12 ) + ϕ(22 ) = ϕ(21 ) + ϕ(22 ) one has equality, by ϕ(2) = 1. Thus all solutions of equation 1) are n = 1 and n = 2. 2) Let d be a divisor of n. Then n = kd, so n − d = (k − 1)d ≥ d if k ≥ 2. This is true, if n ≥ 2, d 6= n. For such d’s one has (σ(d))n−d ≥ (σ(d))d . On the other hand, if n ≥ 3, there is at least a divisor d such that n − d ≥ 2d (since n ≥ 3d, i.e. X Xk ≥ 3 e.g. take d = 1, when n ≥ 3 = 1 · 3). Then clearly n−d (σ(d)) > (σ(d))d in such a case. For n = 1, 2 there is no solution, too. 3) We shall use the following property of the function ψ (see e.g. [1]): ψ(ab) ≥ aψ(b), (∀) a, b ≥ 1.

(2)

Now, if n = dk, then n − d = d(k − 1), so ψ(n − d) ≥ (k − 1)ψ(d) > ψ(d), if k − 1 > 1 for at least a divisor d. Clearly, for n ≥ 3 one has such a divisor, namely d = 1. Thus the equality ψ(d1 ) + · · · + ψ(dr ) = ψ(n − d1 ) + · · · + ψ(n − dr )

(3)

cannot be true. Clearly n ≥ 2, and we do not have solutions.

References [1] J. Sándor, On Dedekind’s arithmetical function, Seminarul de teoria structurilor, no.51, 1988, Univ. of Timi¸soara, 1-15.

15

Equations with composite functions

Let d(n), σ(n), ϕ(n) denote the number of divisors, the sum of divisors, and Euler’s totient, respectively. Our aim is to solve completely the four equations stated in the title. 351

The following well-known inequalities will be used (see e.g. [3]): √ Lemma 1. d(n) < 2 n for all n ≥ 1. (1) Lemma 2. ϕ(n) ≤ n − 1 for all n ≥ 2. (2) σ(n) n+1 Lemma 3. ≤ . (3) d(n) √2 Lemma 4. σ(n) < n n for all n ≥ 2. (4) Theorem 1. All solutions of equation σ(d(n)) = n are n = 1, 3, 4, 12. In fact, σ(d(n)) < n for all n > 12. (5) Proof. By (3), √ p d(n) + 1 2 n+1 σ(d(n)) ≤ d(d(n)) < 2 d(n) 2 2 √ q √ √ 2 n+1 √ √ 8x2 + 8x + 2. Let f (x) = x3 − 8x2 − 8x − 2. Since f 0 (x) = 3x2 − 16x − 8, f 00 (x) = 6x − 16, clearly for all x ≥ 9 we have f 00 (x) > 0 so f 0 (x) > f 0 (9) > 0, f (x) ≥ f (9) = 7 > 0. Thus for n ≥ 81, the inequality σ(d(n)) < n is proved. A computer search shows that this is true for all n > 12. For n = 1, 3, 4, 12 there is equality, finishing the proof of Theorem 1. Remark. For Theorem 1 see also [2]. Theorem 2. All solutions of equation d(σ(n)) = n are n = 1, 2, 3. In fact, d(σ(n)) < n for all n ≥ 4. (6) p √ p 3/4 Proof. By (1) and (4), d(σ(n)) < 2 σ(n) < 2 n n = 2n < n ⇔ 16n3 < n4 ⇔ n > 16. For n ≥ 4, n ≤ 16 a direct computation applies. It is immediate that n = 1, 2, 3 are solutions. Theorem 3. The single solution of ϕ(d(n)) = n is n = 1. In fact, for any n ≥ 2 one has ϕ(d(n)) < n. (7) Proof. By (2) and (1) one can write successively for n ≥ 2, ϕ(d(n)) ≤ √ √ √ n+1 (i.e. ( n − 1)2 > 0). Clearly n = 1 d(n) − 1 < 2 n − 1 < n since n < 2 is the single solution. Theorem 4. The single solution of equation d(ϕ(n)) = n is n = 1. In fact, for all n ≥ 2 one has d(ϕ(n)) < n. (8) p Proof. Let n ≥ 2. Then by (1) and (2) one can write d(ϕ(n)) < 2 ϕ(n) < √ 2 n ≤ n ⇔ n ≥ 4. For n = 2, 3 one has ϕ(2) = 1, ϕ(3) = 2, d(1) = 1 < 2, d(2) = 2 < 3. Remark. For difficulties regarding the equations σ(ϕ(n)) = n and ϕ(σ(n)) = n, see [1], [4]. 352

References [1] R. K. Guy, Unsolved problems in number theory, Third Edition, 2004, Springer Verlag. [2] J. Sándor, On a note by Amarnath Murthy, Octogon Math. Mag. 9(2001), no. 2, 836-838. [3] J. Sándor, Geometric theorems, Diophantine equations, and arithmetic functions, American Research Press, Rehoboth, 2002. [4] J. Sándor, Handbook of number theory, II, Kluwer Acad. Publ., to appear.

16

On d(n) + σ(n) = 2n

In a recent paper, Jason Earls [1] raised some problems and conjectures on abundant and deficient numbers. Particularly, in Problem 17 he denotes the deficiency of n by α(n) = 2n − σ(n), and notes that n = 1, 3, 14, 52, 130, 184, 656, 8648, 12008, 34688, 2118656 are solutions to α(n) = d(n). Here, as usual, d(n) and σ(n) denote the number, resp., sum- of divisors of n. The above equation is clearly equivalent with d(n) + σ(n) = 2n

(1)

Particularly, in [1] it is asked if (1) has infinitely many solutions. Our aim in what follows is to investigate certain properties of solutions of equation (1). Theorem 1. Let p be a prime and α ≥ 1 a positive integer. Then n = pα is a solution to (1) only if p = 2 and α = 1 (i.e. n = 3). Proof. (1) is equivalent to α+1+

pα+1 − 1 = 2pα , p−1

(2)

or after certain elementary transformations to pα (p − 2) = p(α + 1) − (α + 2)

(3)

For p = 2 we get α = 0; contradiction. Thus p ≥ 3. For α = 1 we get p = 3, as a solution to (3). Now, suppose p ≥ 3 and α ≥ 2. Then, by induction upon α it follows immediately that pα > p(α + 1) − (α + 2), 353

(4)

contradicting (3), since p − 2 ≥ 1. Indeed, for α = 2 one gets p2 > 3p − 4, i.e. p(p − 3) > −4, which is trivial. Now, assuming (4), one can write pα+1 > p[p(α + 1) − (α + 2)] > p(α + 2) − (α + 3) iff p2 (α + 1) > 2p(α + 2) − (α + 3), i.e. p[p(α+1)−2(α+2)] > −(α+3). Here p(α+1)−2(α+2) ≥ 3(α+1)−2(α+2) = α − 1 ≥ 1, and all is clear, since −(α + 3) < 0. Theorem 2. Let p be an odd prime. Then n = 2k p is a solution to (1) only if p = 2k+1 + 2k + 1. Proof. Since d(n) = 2(k+1), σ(n) = (2k+1 −1)(p+1), an easy computation yields that (1) is equivalent to 2k+1 + 2k + 2 − p − 1 = 0, i.e. p = 2k+1 + 2k + 1. Remarks. Therefore, for primes of the form p = 2k+1 + 2k + 1

(4)

we get a solution. For k = 1, p = 7 = prime, so n = 14. For k = 2, p = 13 = prime, so n = 52; for k = 3, p = 23 = prime, so n = 184; for k = 4, p = 41 = prime, so n = 656; for k = 7, p = 271 = prime, so n = 27 · 271 = 34688; for k = 10, p = 2069 = prime, so n = 210 · 2069 = 2118656. For k = 13 we get the prime p = 214 + 27 = 16411, which gives the number n = 213 · 1641 = 134448912, not listed by Earls. It seems very likely that there are infinitely many primes of the form (4). If this is true, then clearly an infinite number of solutions to (1) will be provided. Theorem 3. Let p < q be odd primes. Then n = 2k · p · q is a solution to (1) only if (p + q)(2k+1 − 1) + 2k+1 + 4k + 3 = p · q (5) There are no solutions for p = 3. The only solution for p = 5 is n = 2 · 5 · 13 = 130. We cannot have k = 2. For k = 3 the only solutions are p = 19, q = 79 and p = 23, q = 47. Proof. d(n) = 4(k + 1), σ(n) = (2k+1 − 1)(p + 1)(q + 1), and after some elementary computations, (1) can be transformed into (5). For p = 3 remark, that since 2k+1 − 1 ≥ 3, the left side of (5) becomes > 3q, while the right side is 3q; a contradiction. For p = 5 we cannot have k ≥ 2, since then 2k+1 −1 ≥ 7, so the left side of (5) will be greater than the right side. For k = 1, however we can obtain the solution q = 13, yielding n = 130. For k = 2 we get 7(p + q) + 19 = pq, i.e. (p − 7)(q − 7) = 68 (6) Here p − 7 and q − 7 are both even, and (68) is impossible, since 68 = 2 · 34, where p = 9 is not a prime. For k = 3, however we get the equation 15(p + q) + 31 = pq, i.e. (p − 15)(q − 15) = 256 354

(7)

Here p − 15, q − 15 are even, and writing 256 as 256 = 2 · 128 = 4 · 64 = 8 · 32 = 16 · 16, we get the only solutions p − 15 = 4, q − 15 = 64; p − 15 = 8, q − 15 = 32; i.e. p = 19, q = 79, and p = 23, q = 47 (p − 15 = 2, q − 15 = 128 do not give solution, since q = 128 + 15 = 143 = 11 · 13 is not prime). Remarks. Thus for k = 3 the only solutions are n = 23 · 23 · 47 = 8648 and n = 23 · 19 · 79 = 12008, which are listed by J. Earls in [1]. Relation (5) can be written also as (p − 2k+1 + 1)(q − 2k+1 + 1) = 22k+2 − 2k+1 + 4k + 4

(8)

Thus for k = 4 one obtains: (p − 31)(q − 31) = 1012

(9)

Since 1012 = 2 · 506 = 2 · 2 · 253, we have the only possibility p − 31 = 2, q−31 = 506, which do not give a solution, since p = 33 is not prime! Therefore, we cannot have k = 4. Similarly for k = 5 one has (p − 63)(q − 63) = 4056,

(10)

and since 4056 = 2 · 2028 = 4 · 1014 = 8 · 507, we cannot have again solution by p = 63 + 4 = 67, q = 63 + 1014 = 1077, which is not prime.


355

356

Chapter 9

Arithmetic functions ”... There still remain three studies suitable for free man. Arithmetic is one of them.” (Plato)

”... The story was told that the young Dirichlet had as a constant companion all his travels, like a devout man with his prayer book, an old, worn copy of the Disquisitiones Arithmetica of Gauss.” (H. Tietze)

357

1

The non-Lipschitz property of certain arithmetic functions

Let d, σ, ϕ, π be the usual arithmetic functions denoting respectively the number of divisors, sum of divisors, Euler’s totient and the counting function of primes, respectively. We note here that none of these functions has the Lipschitz property, i.e. |f (x) − f (y)| ≤ L|x − y| (L > 0 constant), x, y ∈ N.

(1)

First remark that by Dirichlet’s theorem, for all m ≥ 1 there are infinitely many primes n of the form n = k · 2m − 1. Now d(n + 1) − d(n) = d(k2m ) − 2 ≥ d(2m ) − 2 = m + 1 − 2 = m − 1, so for m → ∞ we get d(n + 1) − d(n) → ∞. Therefore lim sup[d(n + 1) − d(n)] = +∞ n→∞

(2)

which means that d cannot have the Lipschitz property (1). We have used also the property d(ab) ≥ d(a). For the function σ remark that:   1 1 − σ(n!) − 1 σ(n!)  σ(n!) − σ(1) σ(n!)   , = =  1  n! − 1 n! − 1 n! 1− n! and by

σ(n!) X 1 1 1 = ≥ 1 + + ··· + n! d 2 n d|n!

one has

σ(n!) → ∞, n → ∞, giving n! lim

n→∞

σ(n!) − σ(1) = +∞. n! − 1

(3)

This contradicts (1) for x = n!, y = 1, f = σ. For the function ϕ it is well-known that (see e.g. [1]) lim sup[ϕ(n + 1) − ϕ(n)] = +∞ n→∞

so ϕ doesn’t have as well, the Lipschitz property. 358

(4)

Finally, since π(m) > m log m and π(m) < m log m − m log log m for sufficiently large m, one has π(3n) − π(2n) 3n log 3n − 2n log 2n + 2n log log 2n > → ∞ as n → ∞, n n so

· lim

n→∞

¸ π(3n) − π(2n) = +∞. 3n − 2n

(5)

Thus π cannot have the property (1).

References [1] J. Sándor, D.S. Mitrinović (in coop. with B. Crstici), Handbook of number theory, Kluwer Acad. Publ., 1995.

2

On certain open problems considered by Murthy In a recent note A. Murthy [1] has considered the indices Id = Id (n) =

n , d(n)

IS = IS (n) =

n , S(n)

Iσ = Iσ (n) =

n , σ(n)

Iϕ = Iϕ (n) =

n , ϕ(n)

where d(n), S(n), ϕ(n), σ(n) are respectively the number of divisors of n, the Smarandache function, the Euler totient, and the sum of divisors of n. He conjectures that the functions Id , IS , Iϕ , Iσ : N∗ → N∗ are all surjective functions. For example, in case of Iσ this means that every positive integer k is the abundancy index of some positive integer µn, i.e. ¶ there exists n ≥ 1 with σ(n) σ(n) = k. It is well-known that the sequence is dense in (1, ∞), n n n≥2 so that the above question is natural even in the weaker form: is every rational number k the abundancy index Iσ (n) of some integer n? We note that for a fixed k, the solutions of (1) are called k-perfect numbers; the numbers n with n(σ(n)) (i.e. k ∈ N) are called multiperfect numbers. These are extremely difficult problems, and only partial solutions are known, see e.g. R. Laatsch [2] for a presentation of the abundancy index. (See also the recent book by the author [7].) H.J. Kanold showed first that the set of all integers n with n|σ(n) has asymptotic density zero. For similar results on the set of integers with d(n)|n, 359

see [5]. For example, in 1985 C.A. Spiro proved that if S(x) = card{n ≤ x : d(n)|n}, then S(x) = (1 + θ(1))x(log x)−1/2 (log log x)−1

(2)

which improves S(x)/x → 0 as x → ∞. P. Erdös [6] proved that for n 6= 3, 5 one has d(n!)|n! and this yields a weaker version of (2). We now show that Murthy’s conjecture for the surjectivity of Iϕ is not true. Thus the proposition: for all k one can find n such that n =k ϕ(n)

(3)

is not true! Writing (3) as n = ϕ(n)k, and remarking that for n ≥ 3, ϕ(n) is even, this equality implies that for any k, n ≥ 3 is even. When n = 1, then clearly k = 1; for n = 2 one has k = 2. Let n = 2m N (N odd) be even. Then (3) implies 2m N = 2m−1 ϕ(N )k, i.e. 2N = ϕ(N )k. (4) If N = 1, this is possible, namely for k = 2 (i.e. n = 2m are the solution of (3)); let N ≥ 3, and suppose k is even. Then the left side of (4) is divisible only by 2, but the right side by 22 , a contradiction. But the real surprise is that one can prove that all integer solutions (n, k) or (3) are given by n = 2a , a ≥ 0, n = 2a 3b , a, b ≥ 1, when k ∈ {1, 2, 3}. Thus Iϕ (N∗ ) ∩ N∗ = {1, 2, 3}.

(5)

Result (5) is essentially due to W. Sierpinski [3]. (j) For the complete solutions of ϕn |n (ϕ(j) is the j-th iteration) see M. Hausman [4]. For example, when j = 2, all solutions are given by n = 1, 2a , 3, 2a · 3b , 2a · 5, 2a · 7,

a, b ≥ 1

(6)

so one gets easily that Iϕ◦ϕ is not surjective. We now prove that Murthy’s conjecture on the surjectivity of IS , where S is the Smarandache function, is correct. Let k be a given positive integer, and p > k a prime. Remark that S(kp) = p, since p! = 1 · 2 · 3 . . . k . . . p is the least kp kp n factorial such that kp|p!. Now, by = = k one has = k with S(kp) p S(n) n = kp. The fact that Id is not surjective (e.g. 18 6∈ Id (N∗ )) has been proved recently by M. Lee, and independently, by J. Sándor (see [8]). 360

References [1] A. Murthy, A conjecture on d(n) and the divisor function itself as divisor with required justification, Octogon Math. Mag., 11(2003), no.2, 647-650. [2] R. Laatsch, Measuring the abundancy of integers, Math. Mag., 59(1986), no.2, 84-92. [3] W. Sierpinski, Elementary theory of numbers, Warsaw, 1964. [4] M. Hausman, The solution of a special arithmetic equation, Canad. Math. Bull. 25(1982), 114-117. [5] J. Sándor, D.S. Mitrinović (in coop. with B. Crstici), Handbook of number theory, Kluwer Acad. Publ., 1996. [6] P. Erdös, Problem 3 of M. Schweitzer Math. Competition, Mat. Lapok (Budapest), 25(1974), no.3-4(1976), 353-357. [7] J. Sándor (in coop. with B. Crstici), Handbook of number theory, II, Springer Verlag, 2005. [8] J. Sándor, On certain open problems considered by A. Murthy, Octogon Math. Mag., 13(2005), no.1B, 894-896.

3

On an inequality of Moree on d(n)

Let d(n) denote the number of all distinct divisors of n. Then it is easy to see that d(nm) ≥ max{d(n), d(m)} (1) for all n, m ∈ N. This relation has been applied by us in certain problems related to a factorial of a number ([2]). Inequality (1) may be defined as follows: d(nm) ≥ d(n) + d(m) − 1

(2)

for all m, n ∈ N. This is due to P. Moree [1]. Clearly, (2) improves (1) since d(m) − 1 ≥ 0, d(n) − 1 ≥ 0. On the other hand, (2) may be even refined, as for all m > 1, n > 1 one has d(nm) > d(n) + d(m).

(3)

Therefore, for such values of n, m one has in fact d(nm) ≥ d(n) + d(m) + 1. 361

(4)

Let k ≥ 0 be a nonnegative integer. Let σk (n) be the sum of kth powers of divisors of n, i.e. X σk = dk . d|n

We shall extend Moree’s inequality as follows: σk (mn) ≥ σk (m) + σk (n) − 1

(5)

for all m ≥ 1, n ≥ 1, k ≥ 0. We note that for k = 0 this gives relation (2). First remark that if at least one of m, n is = 1, one has equality in (5). So we may suppose m, n > 1. Then the following stronger result will be true: σk (mn) > σk (m) + σk (n) for all m, n > 1.

(6)

If (m, n) = 1 this is trivial, since by the multiplicativity of σk one can write σk (mn) = σk (m)σk (n). But ab > a + b for a, b ≥ 2 since (a − 1)(b − 1) > 1 implies ab > a + b. Let us suppose that (m, n) > 1; let Y Y Y 0Y m= px qy , n = px rz , where (p, q) = (p, q) = (q, r) = 1 and x, x0 , y, z are positive integers (y or z may be zero). We do not use indices for simplicity. Now, ³Y ´ ³Y ´ ³Y ´ 0 σk (mn) = σk px+x σk q y σk rz ³Y ´ ³ ³Y ´ ³Y ´´ 0 ≥ σk px+x σk q y + σk rz ³Y ´ ³Y ´ ³Y ´ 0 0 = σk px+x σk (q y ) + σk px+x σk rz ³Y ´ ³Y ´ Y ³Y 0 ´ ³Y ´ Y 0 > pkx σk px σk qy + pkx σk px σk rz ³Y ´ ³Y ´ ³Y 0 ´ ³Y ´ ≥ σk px σk q y + σk px σk rz = σk (m) + σk (n). have úsed the fact that (6) is valid for (m, n) = 1 (in case of ³YWe Y y q , rz = 1) and the following known fact: σk (AB) ≥ Ak σk (B) for all A, B ≥ 1, k ≥ 0 (with strict inequality for A, B > 1). Corollary. For all m, n > 1 one has σ(mn) > σ(m) + σ(n)

(7)

d(mn) > d(m) + d(n).

(8)

and

Proof. Select k = 1, respectively k = 0 in relation (6). 362

References andor, dated 29th April 2003. [1] P. Moree, e-mail to J. S´ [2] J. Sándor, On values of arithmetical functions at factorials, I, Smarandache Notions J., 10(1999), 87-94.

4

On duals of the Smarandache simple function 1. The Smarandache simple function is defined by Sp (n) = min{k ≥ 1 : pn |k!}

(1)

where p is a fixed prime number. The multiplicative ”dual” of this function has been defined by us (see [1]) as: Sp∗ (n) = max{k ≥ 1 : k!|pn }.

(2)

The additive variants of these functions are Sp (x) = min{k ≥ 1 : px ≤ k!}

(3)

Sp∗ (n) = max{k ≥ 1 : k! ≤ pn }

(4)

where p > 1 is a fixed real number, and x > 0 is a real number. These functions have been studied in [1]. For example, we have proved that:

and that the series

log Sp∗ (x) ∼ log x as x → ∞

(5)

µ ¶ ∞ X 1 log log n α n log Sp∗ (n)

(6)

n=1

is convergent for α > 1, and divergent for α ≤ 1. The additive variants of the Smarandache function S(n) have been introduced in [2], and further generalized by C. Adiga and T. Kim [3] and C. Adiga, T. Kim, D.D. Somashekara and A.N. Fathima [4]. 2. The aim of this note is to consider the function Sp∗ (n) given by (2), which (though introduced in [10] has not been studied up to now). First we note that the following simple result is true: Theorem 1. ½ 1, if p ≥ 3 Sp∗ (n) = = L(p). 2, if p = 2 363

Proof. If p = 2, then we cannot have k ≥ 3, since then 3|k!, but 3 - 2n . Clearly kmax = 2, since 2! = 2|2n for all n ≥ 1. For p ≥ 3 remark that we cannot have k ≥ 2, since then 2|k!|pn would imply 2|pn , impossible since pn is odd. Therefore (7) is proved. Remarks. Therefore, Sp∗ (n) is independent of n, i.e. Sp∗ (1) = Sp∗ (2) = · · · = Sp∗ (n) = constant, when p is fixed. The things are different when one considers Sp∗ (n) given by (2) not only for primes p, but arbitrary positive integers a ≥ 1. Sa∗ (n) = max{k ≥ 1 : k!|an }. For example,

½ S6∗ (n) =

3, if n = 1, 2 4, if n ≥ 3

(8)

(9)

Indeed, remark that k!|6 and k!|36 are valid for k ≤ 3 only, while k!|63 = · 33 is true for k ≤ 4. Now, for k ≥ 5, k! ≡ 0 (mod 10), while clearly 6n 6≡ 0 (mod 10). This proves (9). Remark also that when a is a prime-power, a = ps , then Sps ∗ (n) = So∗ (n) = L(p) for any s, with the same proof as (7). Hence the function f (a) = Sa∗ (n) (n = fixed) is a prime-independent function. Also, for a = odd, f (a) = 1 for any n ≥ 1 (since for k ≥ 2, k! is even). Generally, for all a, a similar situation to (9) is true; namely Sa∗ (n) can take a finite number of values. Theorem 2. Let a = pα1 1 pα2 2 . . . pαr r be the prime factorization of a > 1 and suppose p1 < p2 < · · · < pr . Then if p 6∈ {p1 , . . . , pr } is any prime, then 23

Sa∗ (n) ≤ p − 1.

(10)

Proof. Suppose that k = Sa∗ (n) ≥ p. Then p!|k!|an implies that p!|an , so p|an , impossible by the definition of p. Clearly, this is true for any prime p 6∈ {p1 , . . . , pr }. For example, for a = 6 = 2 · 3 one can select p = 5 and (10) gives S6∗ (n) ≤ 4, in concordance with (9). For a = 30 = 2 · 3 · 5 one has S30∗ (n) ≤ 6.

(11)

Remark that for a = odd one can select p = 2, so Sa∗ (n) ≤ 1, which by Sa∗ (n) ≥ 1 gives the relation Sa∗ (n) = 1, i.e. f (a) = 1. Corollary. If (a, 3) = 1, then Sa∗ (n) ≤ 2. 364

(12)

For example, S10∗ (n) = 2 for any n ≥ 1. 3. As we have seen, the function f (a) = Sa∗ (n) for any fixed n ≥ 1, has the property that f (ps ) = f (p) = L(p), for any s ≥ 1, i.e. f is primeindependent. Now, one can ask what are the functions f , if we suppose that f is multiplicative? Theorem 3. Let f be multiplicative, prime-independent, and f (p) = L(p) for primes p (where L is given by (7)), f (1) = 1. Then one has ½ 1, if n is odd f (n) = (13) 2, if n is even Y Proof. Let 1 < n = pα be the prime factorization of n. If n is odd, then clearly Y Y Y f (n) = f (pα ) = f (p) = L(p) = 1 for p ≥ 3. Now, if n is even, n = 2a

Y

pα , then

f (n) = f (2a )

Y

f (pα ) = L(2) · 1 = 2

and (13) is proved. Remark. (13) gives an example of a strongly-multiplicative function (i.e. multiplicative and prime-independent). Now, search for additive functions, which are prime-independent (so, they will be strongly-additive) and which satisfy (7). Theorem 4. Let g be an additive, prime independent function such that g(p) = L(p) for primes p, and let g(1) = 0. Then  1, if n is an odd prime power      2, if n is a power of 2 ω(n), if n is odd, and ω(n) ≥ 2, (14) g(n) =   ω(n) + 1, if n is even, but it has    at least an odd divisor, where ω(n) denotes the number of distinct prime divisors of n. Proof. Assume g(ab) = g(a)+g(b) for any (a, b) = 1, g(pα ) = g(p) = L(p). Then if nY = pα (p odd), g(n) = L(p) = 1 for p ≥ 3. If n = 2α , clearly g(n) = 2. Let n = pα with π(n) ≥ 2, p ≥ 3. Then g(n) =

X

g(pα ) =

X

g(p) = 365

X

L(p) =

X

1 = ω(n),

if all p is odd. If n = 2a

Y

pα , then

g(n) = g(2a ) + = L(2) +

X

X

g(pα ) = g(2) +

X

g(p)

L(p) = 2(ω(n) − 1) = ω(n) + 1.

Together with g(1) = 0, (14) gives an example for a strongly-additive function.

References [1] J. Sándor, On additive analogues of certain arithmetic functions, Smarandache Notions J., 14(2004), 128-133. [2] J. Sándor, On an additive analogues of the function S, Notes Number Theory Discr. Math., 7(2001), 91-95. andor’s function, Proc. Jang[3] C. Adiga, T. Kim, On a generalization of S´ jeon Math. Soc., 5(2002), 121-124. [4] C. Adiga, T. Kim, D.D. Somashekara, A.N. Fathima, On a q-analogue of S´ andor’s functions, JIPAM, 4(2003), no.5, article 84.

5

A modification of the Smarandache function

1. For a given function f : N∗ → N∗ and a given set A ⊂ N∗ , in papers [1], [2] we have introduced the arithmetical function FfA (n) = min{k ∈ A : n|f (k)}

(1)

if this is well defined. We have also considered the ”dual” function GA g (n) = max{k ∈ A : f (k)|n},

(2)

if this is again, well defined. In various papers (see [3] and [4]), which have been recently published, or are under publication, we have studied certain properties of these functions for A = N ∗ and f (k) = g(k) = k!,

f (k) = g(k) = ϕ(k),

f (k) = g(k) = σ(k), f (k) = g(k) = S(k),

f (k) = d(k),

f (k) = g(k) = T (k),

366

etc., where ϕ, σ, d, S, T represents Euler’s totient, the sum of divisors, number of divisors, Smarandache’s function, and the product of divisors functions, respectively. For additive analogs, see e.g. [5], [6], [7]. 2. The aim of this note is the introductions and preliminary study of a particular case of (1) when A = N ∗ and f (k) = k! + (k − 1)! = (k − 1)!(k + 1). Here 0! = 1. Let us denote this function by F (n) = min{k ≥ 1 : n|(k − 1)!(k + 1)}.

(3)

This seems to be closely analogous to the Smarandache function S(n) = min{k ≥ 1 : n|k!},

(4)

and we shall call it as ”a modification of the Smarandache function”. 3. In spite of the similarity between (3) and (4), the two functions S(n) and F (n) have quite distinct properties. For example, it is well-known, that S(p) = p for all primes p. For the function F (p) we have: Theorem 1. F (p) = p − 1 for all primes p.

(5)

(6)

Proof. Clearly, p|(p − 2)!p for all primes p, so F (p) ≤ p − 1. Incidentally, for general n ≥ 2 we have n|(n − 2)!n, so by (3) we have: F (n) ≤ n − 1 for all n ≥ 2.

(7)

On the other hand, if k ≤ p − 2, then p - (k − 1)!(k + 1), since k − 1 ≤ p − 3 and k + 1 ≤ p − 1, so the prime factors of (k − 1)! and k + 1 are less than p. This proves relation (6). It is well-known and obvious that: S(k!) = k, for all k.

(8)

Theorem 2. F (k!) = k + 1 for all k ≥ 3,

367

F (1!) = F (2!) = 1.

(9)

Proof. Clearly F (1) = F (2) = 1 since 1|0! · 2, 2|0! · 2, where 0! = 1 by the known convention. Let now k ≥ 3. Clearly k!|k!(k + 2), so F (k!) ≤ k + 1. Let us suppose that there is an m ≤ k with k!|(m − 1)!(m + 1). Since m ≤ k, so m − 1 ≤ k − 1, thus (k − 1)! = (n − 1)!l, l ≥ 1, and on the other hand, one has (m − 1)!|(m + 1) = k!A. Since k! = (k − 1)k, one gets m + 1 = klA. Thus m + 1 ≥ k and by m + 1 ≤ k + 1, we can have only m + 1 = k or m + 1 = k + 1. Since (k, k + 1) = 1, this last possibility cannot hold. But for m + 1 = k we get l = A = 1, so k − 1 = m − 1, i.e. k = m, impossible by m!|(m − 1)!(m + 1). This proves relation (9). Corollary 1. For infinitely many n one has S(n) < F (n).

(10)

Corollary 2. For infinitely many m one has S(m) > F (m).

(11)

Proof. Put n = p in (11). Then S(p) = p > p − 1 = F (p) by (5) and (6). Now, let n = k!, k ≥ 3 in (10). Then S(k!) = k < k + 1 = F (k!) by (8) and (9). Remark. Since F (n) ≥ 1, by (7) we get the limit p lim n F (n) = 1. (12) n→∞

References [1] J. Sándor, On certain generalizations of the Smarandache functions, Notes Numb. Th. Discr. Math., 5(1999), no.2, 41-51. [2] J. Sándor, On certain generalizations of the Smarandache functions, Smarandache Notions J., 11(2000), 202-212. [3] J. Sándor, The product of divisors minimum and maximum functions, RGMIA Research Report collection, 7(2004), no.2, art.18, 11. [4] J. Sándor, A note on the divisor minimum function, Octogon Math. Mag., 12(2004), no.1, 273-275. [5] J. Sándor, On an additive analogue of the function S, Notes Number Theory Discr. Math., 7(2001), 91-95. 368

[6] J. Sándor, On additive analogues of the function S, Smarandache Notions J., 14(2004), 128-133. [7] J. Sándor, An additive analogue of the Euler minimum function, Adv. Stud. Contemp. Math., 10(2005), no.1, 53-62.

6

The star function of an arithmetic function

Let d(n) be the number of distinct positive divisors of n. Recently, in an interesting note, Murthy and Bencze [1] have considered a ”star function” X d∗ (n) = d(k), k|n

and studied some of its properties, as well as analogous notions. We must note here that the ”star” notation is already a standard notation when considering ”unitary divisors” and associated arithmetic functions³ (seeé.g. [2], [3]). An n integer k is called a unitary divisor of n if k|n and k, = 1. Then the k number of unitary divisors of n is denoted in the literature as X 1. d∗ (n) = n k|n,(k, k )=1 Therefore, when using this ”star” notation, we always must note that this is not the unitary divisor theory notation. The star function of an arbitrary function f : N → R can be defined similarly as X f ∗ (n) = f (k). k|n

Thus we get: Theorem 1. If f is multiplicative (i.e. f (mn) = f (m)f (n), for all (m, n) = 1 and f (n) 6= 0), then f ∗ is multiplicative, too, and for the prime factorization n = pα1 1 . . . pαr r one has r Y f (n) = (1 + f (pi ) + · · · + f (pαi i )) ∗

(1)

i=1

Particularly, d∗ (n) =

r Y (α + 1)(α + 2) Y (α1 + 1)(α1 + 2) = . 2 2 α i=1

p kn

369

(2)

Proof. The first part is a classical result, see e.g. [4]. Clearly f (1) = 1. Since d(1) + d(p) + · · · + d(pα ) = 1 + 2 + · · · + (α + 1) =

(α + 1)(α + 2) , 2

the second part also follows. Remark 1. Let ϕ be Euler’s totient. By Gauss theorem (which easily follows from (1), too) ϕ∗ (n) = n, so the star function of Euler’s totient is the identity function. If σ denotes the sum of divisors function, then, by (1) σ ∗ (n) =

r Y [1 + (pi + 1) + · · · + (pαi i + · · · + 1)]. i=1

This must not be confused with the sum of unitary divisors function, for which r Y σ ∗ (n) = (1 + pαi i ) i=1

(see e.g. [3], [5]). Theorem 2. Let ω(n), resp. Ω(n) denote the number of distinct, respectively total-number of prime factors of n. Then 1 · ¸ i · d∗ (n) ¸ ω(n) 1 1h 1 Ω(n) 3 ω(n) ≤ ≤ 2+ . (3) 1 + d(n) ≤ 2 2 d(n) 2 ω(n) Proof. By (2) one can write d∗ (n) =

d(n) (α1 + 2) . . . (αr + 2). 2ω(n)

Now, by the arithmetic-geometric inequality one has µ ¶ µ ¶ α1 + 2 + · · · + αr + 2 r Ω(n) ω(n) (α1 + 2) . . . (αr + 2) ≤ = 2+ , r ω(n) since ω(n) = r and Ω(n) = α1 + · · · + αr . This gives the right side of (3). For the second part of left side apply Chrystal’s inequality p √ r (1 + x1 ) . . . (1 + xr ) ≥ r x1 . . . xr + 1 with xi = αi + 1, i = 1, r. Since d(n) = (α1 + 1) . . . (αr + 1) ≥ 2r , 370

the first inequality of (3) follows. Corollary 1. The normal order of magnitude of the arithmetical function µ F =

d∗ d

¶1

ω

3 . 2 Proof. This follows by (3) and the classical result due to Hardy and RaΩ(n) manujan that the normal order of magnitude of is 1 (see e.g. [6]). ω(n) Theorem 3. Let d∗∗ be the star function of d∗ . Then is

d∗∗ (n) ≤ d∗ (n)d(n). Proof. FirstY we note the property m|n ⇒ d∗ (m) ≤ d∗ (n). Indeed if Y n= pa , m = pb with b ≤ a, then d∗ (m) =

Y (b + 1)(b + 2) 2

≤

Y (a + 1)(a + 2) 2

= d∗ (n).

Now, by definition, X X X d∗∗ (n) = d∗ (k) ≤ d∗ (n) = d∗ (n) 1 = d∗ (n)d(n) k|n

k|n

k|n

by the above proved property. Remark 2. The notation d∗∗ (n) should not be confused with the number of ”bi-unitary” divisors of n (see [4], [6]), with the same notation. Theorem 4. σ ∗ (n) ≤ σ(n)d(n). Proof. Similarly as above m|n ⇒ σ(n) ≤ σ(n), so X σ ∗ (n) = σ(k) ≤ σ(n)d(n). k|n

References [1] A. Murthy, M. Bencze, Extending the scope of some number theoretic functions, Octogon Math. Mag., 11(2003), no.1, 110-113. [2] E. Cohen, The number of unitary divisors of an integer, Amer. Math. Monthly, 67(1960), 879-880. 371

[3] E. Cohen, Arithmetical functions associated with the unitary divisors of an integer, Math. Z., 74(1960), 66-80. [4] J.M. DeKoninck, A. Ivić, Topics in arithmetical functions, North Holland Math. Studies, 72(1980). [5] J. Sándor, L. Tóth, On certain number-theoretic inequalities, Fib. Quart., 28(1990), 255-258. [6] J. Sándor, D.S. Mitrinović, Handbook of number theory, Kluwer Acad. Publ., 1995. [7] J. Sándor, On the arithmetical functions dk (n) and d∗k (n), Portugaliae Math., 53(1996), 107-115.

7

On Jordan’s arithmetical function

Prelimiaries Jordan’s arithmetical function is X a generalization of Euler’s totient func1 = the number of all k-tuples tion ϕ. By definition, ϕk (n) = (a1 ,...,ak ,n)=1 1≤a1 ,...,ak ≤n

(a1 , . . . , ak ) with all components between 1 and n such that (a1 , . . . , ak , n) = 1. We note that this function has some applications in the theory of linear groups. For k = 1 we have ϕ1 = ϕ ([4]). Furthermore, let σk (n) be the sum of kth powers of divisiors of n, i.e. X k σk (n) = d . For k = 1 and k = 0 we reobtain the well-known arithmetical d|n

functions σ1 (n) = σ(n), the sum of divisors of n, and σ0 (n) = d(n), the number of divisors of n. The aim of this note is to prove some interesting relations for the above mentioned arithmetical functions. First we need some lemmas. Lemma 1. X ϕk (d) = nk d|n

Proof. Let us consider the set S = {(a1 , . . . , ak ) : a1 , . . . , ak ∈ N∗ , 1 ≤ a1 , . . . , ak ≤ n} =

[ d|n

where (a1 , . . . , ak , n) = d iff d|n. 372

Sd ,

³ n´ n From a1 = b1 d, . . . , ak = bk d, b1 , . . . , bk , = 1 we get b1 ≤ , . . . , bk ≤ d d ³n´ n and so Sd has ϕk distinct elements, the set S trivially has a number of d d k n elements, hence we obtain X ³n´ = nk . ϕk d d|n

n Since |n iff d|n, we have the result. d Lemma 2. X µ(d) ϕk (n) = nk dk d|n

where µ denotes the M¨ obius arithmetical function. Proof. Apply Lemma 1 and the Möbius inversion formula ([2], [3]). ϕr (m) ϕr (n) Lemma 3. If m|n (m divides n) then ≥ . r m nr Proof. It is immediate from the definition (and Lemma 2) that ϕr (ab) ≤ (ϕr (b))a, so writing n = qm, and applying this inequality the result follows. Theorem 1.³ ´ X n ϕk (i)d a) = σk (n), n ≥ 1 i i|n ³n´ X b) ϕk (i)σk = nk d(n), n ≥ 1. i i|n

Proof. It is easy to see that ϕk and σk , d are multiplicative functions and by a known result ([2], [3]) the left sides in a) and b) are also multiplicative. Thus it is sufficient to prove these relations for n = pa (prime powers). However we shall use another argument, which is based on Dirichlet series. Denote D(f, s) =

∞ X f (n) n=1

ns

the Dirichlet series of the arithmetical function X f . It ³isn ´well-known that D(f ∗ g, s) = D(f, s)D(g, s), where (f ∗ g)(n) = f (i)g is the Dirichlet i i|n

product of f and g, by supposing that the considered series are absolute convergent ([3]). Let Ek be the function Ek (n) = nk and U (n) ≡ 1 the identity function. Then evidently, ³n´ X σk (n) = Ek (i)U . d i|n

373

Since D(Ek , s) =

∞ X n=1

1 ns−k

= ζ(s − k) (which is true for Re s > k + 1) and

D(U, s) = ζ(s), the Riemann zeta function, hence we have D(σk , s) = ζ(s − k)ζ(s). For k = 0 one has D(d, s) = ζ 2 (s). Similarly, from Lemma 2, we obtain D(ϕk , s) =

ζ(s − k) . ζ(s)

We now consider the identity ζ(s − k) 2 ζ (s) = ζ(s − k)ζ(s) ζ(s) or D(ϕk , s)D(d, s) = D(σk , s), i.e. D(ϕk ∗ d, s) = D(σk , s). But it is well-known by the uniqueness theorem of Dirichlet series ([8]) that this implies ϕk ∗ d = σk , which is exactly a). In order to prove the second relation, we may consider the identity ζ(s − k) ζ(s − k)ζ(s) = ζ 2 (s − k) ζ(s) or D(ϕk ∗ σk , s) = D(nk d, s), yielding relation b). Next we shall prove: Theorem 2. ∞ X n=1

∞

X xn xn nk for |x| < 1 = ϕk (n) 1 − xn n=1

Proof. We need the following result: X If g(n) = f (d), then d|n ∞ X n=1

n

g(n)x =

∞ X i=1

374

f (i)

xi , 1 − xi

if the involved series are convergent (see [5]). To prove this, let us observe that   ∞ ∞ ∞ X ∞ X X X X  f (i)xij g(n)xn = f (i) xn = n=1

n=1

=

∞ X

i=1 j=1

i|n

f (i)

i=1

xi , for |x| < 1. 1 − xi

Apply now this theorem for f (i) = ϕk (i) and use Lemma 1. One obtains ∞

∞ X

ϕk (i)

X xi = nk xn , i 1−x n=1

i=1

which for k = 1 gives ∞ X

∞

ϕ(i)

X xi x = nxx = . 1−x (1 − x)2 n=1

i=1

By differentiation we have ∞ X

n2 xn =

n=1

hence

∞ X

ϕ1 (x)

i=1

x(1 + x) , (1 − x)3

xi x(1 + x) = . i 1−x (1 − x)3

Remark. The above proved identities imply the interesting relations µ ¶ 1 x= : 2

∞ X ϕ(n) =2 2n − 1

and

n=1

∞ X ϕ1 (n) = 6. 2n − 1

n=1

We conclude with a result on the composite function ϕ(σk (n)). Theorem 3. Let k be an odd natural number. Then we have lim inf n→∞

ϕ(σk (n)) = 0. n

Proof. Let p be a prime number of the form p ≡ −1 (mod p1 . . . ps ), where pi denotes the ith prime number. (By the well-known Dirichlet theorem 375

on arithmetical progressions, there exist such primes ([2], [3]). Then σk (p) = pk + 1 ≡ 0 (mod p1 . . . ps ) and Lemma 3 (r = 1) implies the inequality ¶ s µ ϕ(σk (p)) ϕ(p1 . . . ps ) Y 1 . ≤ = 1− p p1 . . . ps pi i=1

This last product tends to 0 as s → ∞ (see [2], [3]), so the theorem is proved. Remark. For k = 1 Theorem 3 gives a result of L. Alaoglu and P. Erdös [1]. See also [6], [7]. Our argument is completely different. By more complicated argument we can prove that lim inf ϕ((σk (n)) log log log n)/n < ∞. n→∞

References [1] L. Alaoglu, P. Erdös, A conjecture in elementary theory of numbers, Bull. Amer. Math. Soc., 50(1944), 881-882. [2] I. Creang˘a ¸si colectiv, Introducere ˆın teoria numerelor, Ed. Did. Ped., Bucure¸sti, 1965. [3] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, Oxford, 1938. [4] C. Jordan, Traité de substitutions et des équations algébriques, Paris, 1957. [5] G. Pólya, G. Szegö, Aufgaben und Lehrs¨ atze aus der Analysis, Springer Verlag, 1924. [6] J. Sándor, On the arithmetical functions ϕk and σk , Math. Student, 58(1990), 49-54. [7] J. Sándor, On Euler’s arithmetical function, Proc. VIIth National Conf. on Algebra, Bra¸sov, Romania, 1988. [8] E.C. Titchmarsch, The theory of functions, Oxford Univ. Press, 2nd ed., 1978, Theorem 9.6. 376

8

Generalization of a theorem of Lucas on Euler’s totient

Let ϕ be the Euler totient function. In 1845 E. Prouchet [1] proved that for all m, n ≥ 1 one has ϕ(mn) = ϕ(m)ϕ(n)

d , ϕ(d)

(1)

where d = (m, n) = gcd of m and n. Clearly (1) implies the multiplicative property of ϕ, namely ϕ(mn) = ϕ(m)ϕ(n) (2) if (m, n) = 1 and reciprocally, (2) holds only if (m, n) = 1 (since (1) implies ϕ(d) = d, which is true for d = 1). In 1891 E. Lucas [2] proved that for all m, n ≥ 1 ϕ(mn) = (m, n)ϕ([m, n])

(3)

where [m, n] = lcm of m and n. Let f : N∗ → R be an arithmetical function, such that f (n) 6= 0 for all n ≥ 1. Then that relation ϕ(mn) = (m, n)f (|m, n|)

(4)

is satisfied by the following two functions: f1 (n) = n, n ≥ 1,

f2 (n) = ϕ(n).

Indeed, (4) for f2 is exactly Lucas’ theorem (4), while for f1 , (4) is exactly the known relation mn = (m, n)[m, n],

(m, n ≥ 1).

(5)

In 1950 H.N. Shapiro [3], as an extension of Prouchet’s theorem (1) introduced the so-called over-multiplicative functions as follows: f : N∗ → R is called over-multiplicative, if there exists a function g : N∗ → R, g(n) 6= 0 for all n, such that f (mn) = f (m)f (n)g(d), m, n ≥ 1 (6) where d = (m, n). Now, we shall prove the following generalization of the Lucas theorem: 377

Theorem 1. If f is over-multiplicative, then there exists g : N∗ → R, g(n) 6= 0 for all n ≥ 1 such that f (mn) = f (d)g(d)f [m, n],

m, n ≥ 1.

(7)

Proof. Let m = dm0 , n = dn0 , where d = (m, n), so (m0 , n0 ) = 1. Then since (d, dm0 n0 ) = d, by (6) one can write f (d2 m0 n0 ) = f (d)f (dm0 n0 )g(d). But mn = d2 m0 n0 and [m, n] = dm0 n0 by (5). Thus (7) follows. Remarks. If f is over-multiplicative and g(1) = 1, then f is multiplicative. Indeed, letting d = 1 in (6), it follows: f (mn) = f (m)f (m) for (m, n) = 1

(8)

i.e. f is a multiplicative function. Generally speaking, (6) implies only that f is quasi-multiplicative, i.e. f (mn) = kf (m)f (n) for (m, n) = 1

(9)

where k = g(1). Now, relation (7) shows that for over-multiplicative functions f , (4) is true if and only if f (d)g(d) = d, i.e. g(n) =

n , f (n)

n ≥ 1.

(10)

In this case, (6) becomes f (mn) = f (m)f (n)

d f (d)

(11)

generalizing Prouchet’s relation (1). If f (1) = 1, then (11) clearly implies that f is multiplicative, i.e. has the property (8). But not all multiplicative functions satisfy (11)! Indeed, letting m = n in (11) it follows f (m2 ) = mf (m).

(12)

For example, the σ-function (sum of divisors) is multiplicative, but doesn’t satisfy (12). Indeed, e.g. σ(4) = 7 6= 2σ(2) = 6. By induction easily follows that if f (1) = 1, then f (mα ) = mf (mα−1 ) for all m ≥ 1, α ≥ 2. 378

(13)

Now, we prove that, reciprocally, if f is multiplicative and has property (13), then it has also the property (11). Even a stronger result is true: Theorem 2. Suppose that f (1) = 1, f is multiplicative, and f (pα ) = pf (pα−1 ) for all primes Yp≥ Y2, α ≥ 2. Then Y (11) is true. α β α Proof. Let m = p q ,n=p rγ be the canonical factorizations Y Y Y 0 of m and n. Then nm = pα+α qβ rγ , where α+α0 ≥ 2 if (m, n) 6= 1 (if 0

0

(m, n) = 1, then (11) is trivially satisfied). We have f (pα+α ) = pf (pα+α −1 ) = 0 pα+α −1 f (p), since from f (pn ) = pf (pn−1 ) it follows by induction that f (pn ) = pn−1 f (p). Similarly, for all α0 ≥ 2, β0 ≥ 1, γ0 ≥ 2, f (pα0 ) = pα0 −1 f (p), fY (q β0 ) =Yq β0 −1 f (q), f (rγ0 ) = rγ0 −1 f (r). When all α, β = 1, then f (m) = f (p) f (q), f being multiplicative. Therefore, we may assume that all 0 α, β, α , γ ≥ 2. Then f (mn) = f (m)f (n) Y Y Y Y Y Y 0 pα+α −1 q β−1 rγ−1 f (p) f (q) f (r) Y Y 0 Y Y Y Y Y =Y pα−1 q β−1 pα −1 rγ−1 f (p) f (q) f (p) f (r) =

Y p d = , f (p) f (d)

Y 0 as now d = pmin{α,α } and in case min{α, α0 } ≥ 2 one can again apply f (pn ) = pn−1 f (p).

References [1] E. Prouchet, Nouv. Ann. Math., 4(1845), 75-80. [2] E. Lucas, Théorie des nombres I, Gauthier-Villars, Paris, 1891. [3] H.N. Shapiro, On the iterates of certain class of arithmetic functions, Comm. Pure Applied Math., 3(1950), 259-272.

9

Some arithmetic inequalities connected with the divisors of an integer

1. Let 1 = d1 < d2 < · · · < dk = n be the consecutive divisors of n > 1. n n n Then, since di |n we have also |n, and in fact, if di < dj , then > . For di di dj 379

example, let k = 2n (even). Then the divisors in increasing order are 1 = d1 < d2 < · · · < dm < thus dm+1 =

n n n n < < ··· < < =n dm dm−1 d2 d1

(1)

n n n n , dm+2 = , . . . , dk−1 = , dk = d2n = . dm dm−1 d2 d1

In fact, this happens when n is not a square, since it is well-known that if n = pα1 1 . . . pαr r is the prime factorization of n, then k = d(n) = (1 + α1 ) . . . (1 + αr ) and this is even only if all αi are not even, so n is not a square. When n is a square, i.e. n = s2 , then k is odd, k = 2m + 1, and one can write 1 = d1 < d2 < · · · < dm < s < where dm+1 = s, dm+2 =

n n n n < < ··· < < =n dm dm−1 d2 d1

(2)

n n n , . . . , d2m = , d2m+1 = . dm d2 d1

From (1) it is clear that dm ≥ m, so by dm dm+1 = n we have n ≥ m(m + √ 1) > m2 , i.e. m < n. Since k = 2m, one gets √ (3) k = d(n) < 2 n for n 6= square. √ √ In (2) dm+1 = s = n ≥ n + 1, so m ≤ n − 1, implying k = 2m + 1 ≤ √ 2 n − 1, i.e. √ k = d(n) ≤ 2 n − 1 for n = square. (4) We note that actually from m(m + 1) ≤ n one has the slightly stronger relation, improving (3): √ 1 + 4n − 1 k = d(n) ≤ , for n 6= square. (30 ) 2 2. From (1) it follows that dm dm+1 = n, so one can write di di+1 < n for 1 ≤ i ≤ m − 1. Therefore, S = d1 d2 + d2 d3 + · · · + dm−1 dm + dm dm+1 + dm+1 dm+2 + · · · + dk−1 dk =n+

m−1 Xµ i=1

n2 di di+1 + di di+1 380

¶ .

Remark that i(i + 1) ≤ di di+1 < n for 1 ≤ i ≤ m − 1, so S < n + (m − 1)n +

m−1 X i=1

µ ¶ n2 1 2 = mn + n 1 − . i(i + 1) n

Thus S < n2 + mn −

n2 , m

n 6= square.

(5)

n2 Since, by (3) m2 < n, mn − < 0 relation (5) implies the weaker inm equality S < n2 . (6) For n = square (i.e. case (2)) a similar argument gives (6), so the p = 2 case of OQ.1284 [1] is settled. The case p = 3 can be studied in a similar manner, with the difference that e.g. in case (1). S 0 = d1 d2 d3 + d2 d3 d4 + · · · + dm−2 dm−1 dm + dm−1 dm dm+1 +dm dm+1 dm+2 + · · · + dk−2 dk−1 dk µ ¶ µ ¶ m−2 X n3 n2 di di+1 di+2 + = + ndm−1 + , di di+1 di+2 dm−1 i=1

since here dm−1 dm dm+1 = dm−1 dm and dm dm+1 dm+2 = dm

n = ndm−1 ≤ n2 dm

n n n2 · = ≤ n2 . dm dm−1 dm−1

Now di di+1 di+2 ≤ dm−1 dm dm+1 ≤ n2 for 1 ≤ i ≤ m − 2 and di di+1 di+2 ≥ i(i + 1)(i + 2). Using

1 1 1 1 1 1 = · − + · , i(i + 1)(i + 2) 2 i i+1 2 i+2

by addition one gets the well-known identity m−1 X i=1

1 1 1 1 1 = + − < , i(i + 1)(i + 2) 4 2m 2m − 2 4 381

so S 0 < (m − 2)n2 + 2n2 + n3 · i.e.

√ √ 1 n3 3 ≤ n2 n + < n3 ⇔ n2 3 < n3 , 4 4 4

√ √ 3 n < n, i.e. 3 n > 4, which is true for all n ≥ 2, finishing the proof of 4 S 0 < mn2 +

√ n3 n3 ≤ n2 n + < n3 , 4 4

n 6= square

(7)

which improves the case p = 3 of [1]. When n is a square, a similar study applies. I think that the most general case with d1 d2 . . . dp . . . can be performed in the same lines, and I so propose for the patient reader to carry out the calculations. √ We finish with the remark that the inequality d(n) ≤ 2 n sometimes is called as Sierpinski’s inequality [2]. (3), (4), (3’) give improvements of this relation.

References [1] M. Bencze, OQ.1284, Octogon Math. Mag., 11(2003), no.2, 851. [2] W. Sierpinski, Elementary theory of numbers, Warsawa, 1964.

10

A new arithmetic function 1. Let n =

r Y

pai i ≥ 2 be the prime factorization of the positive integer n

i=1

(pi primes, ai ≥ 1). Let us define fk (n) =

¶a r µ k Y p +1 i i

i=1

2

,

k ≥ fixed.

(1)

Let fk (1) = 1 by definition. This arithmetic function appears naturally from a result in [10], where it is proved that ¶a r µ k Y p +1 i i

i=1

2

n

≤

i +1 σk (n) Y pka i ≤ . d(n) 2

i=1

See also [11] for similar relations. 382

(2)

Here σk (n) and d(n) = σ0 (n) stand for the sum of kth powers of divisors of n, and the number of divisors of n, respectively. Let σk∗ (n) and d∗ (n) denote the sum of kth powers of unitary divisors of n, and the number of unitary divisors of n, resp. (see e.g. [2], [6], [9]). Then (2) can be rewritten as fk (n) ≤

σ ∗ (n) σk (n) ≤ ∗k . d(n) d (n)

(3)

There is equality at each side only when n is squarefree. For k = 1 we get the arithmetical function ¶ r µ Y pi + 1 a i f (n) = , r ≥ 1, (4) 2 i=1

with f (1) = 1. Clearly fk (n) and f (n) are examples of multiplicative functions, i.e. they satisfy the functional equation g(nm) = g(n)g(m) for (n, m) = 1. 2. Results on composite functions such as σ(ϕ(n)), d(ϕ(n)), etc., where ϕ(n) is Euler’s totient, are very difficult to obtain. For example, it is conjectured (by Makowski and Schinzel, see e.g. [12] for recent and/or related results) that σ(ϕ(n)) ≥ n/2 (5) for all n. We now prove the following simple result: Theorem 1. For all n ≥ 1 one has f (ϕ(n)) ≤ n/2,

(6)

with equality only for n = 1, 2, 3. Proof. It is known (see [9]) that σk∗ (n) nk + 1 ≤ . d∗ (n) 2

(7)

Now, by (3) and (7) we obtain f (n) ≤

n+1 , implying f (ϕ(n)) ≤ 2

ϕ(n) + 1 n ≤ , since ϕ(n) ≤ n − 1 for all n ≥ 2. An immediate verifica2 2 tion shows that for n = 1, 2, 3 there is equality, and that there are no other such values of n. Remarks. 1) More generally, if an arithmetic function k(n) satisfies n σ(k(n)) ≤ , d(k(n)) 2 383

n ∈ S ⊂ N,

then f (k(n)) ≤ 2) Since for n even ϕ(n) ≤

n for all n ∈ S. 2

n , by the proof of Theorem 1 one can see that 2 n+2 n < for n = even. 4 2

f (ϕ(n)) ≤

(8)

µ

¶ x+1 a xa + 1 ≥ , x > 0, a ≥ 1, applied to x = pki , 2 2a a = ai , after term-by-term multiplication implies 3. The inequality

fk (n) ≥

σk∗ (n) , D∗ (n)

(9)

where D∗ (n) = 2Ω(n) , defined in analogy with d∗ (n) = 2ω(n) , where ω(n), Ω(n) denote the distinct, resp. total number of prime factors of n. Relation (3) and (9) give at once σ ∗ (n) ω(n) log 2 ≤ log k ≤ Ω(n) log 2. (10) fk (n) Theorem 2. The normal order of magnitude of the arithmetical function σ ∗ (n) log k is (log 2) log log n. fk (n) Proof. This follows by the double-inequality (10) and the well-known fact, due to Hardy and Ramanujan, that the normal order of magnitude of ω(n) and Ω(n) is log log n (see e.g. [1], [3]). Theorem 3. σ ∗ (n) log log n lim sup log k ≥ log 2, (11) log n fk (n) n→∞ lim sup m→∞

σ ∗ (n) 1 log k ≤ 1. log n fk (n)

(12)

Proof. This follows by (10) and the known facts lim sup n→∞

ω(n) log log n = 1, log n

lim sup n→∞

Ω(n) log 2 = 1. log n

Theorem 4. Y σ ∗ (n) k = 2x log log x+O(x) as x → ∞, fk (n)

n≤x

384

(13)

X

1 x 1 ∼ · , ∗ σ (n) log 2 log log x 2≤n≤x log k fk (n)

x → ∞.

(14)

Proof. The proof of (13) is based on inequalities (10) and the known facts that (see e.g. [1], [3]) X X ω(n) ∼ x log log x, Ω(n) ∼ x log log x as x → ∞, n≤x

n≤x

while (14) is a consequence of (10) and X 2≤n≤x

1 x ∼ ω(n) log log x

X

and

2≤n≤x

1 x ∼ , Ω(n) log log x

x → ∞,

see e.g. [4]. 4. By (3) and (7) we can write fk (n)/nk ≤ (nk + 1)/2nk , and since for a prime n = p we have fk (p)/pk ≤ (pk + 1)/2pk , clearly lim sup n→∞

1 fk (n) = . 2 nk

(15)

On the other hand, fk (2m )/2mk = (1 + 2−k )m /2m → 0 as m → ∞, thus fk (n) = 0. nk

(16)

σ ∗ (n) σ ∗ (n) 1 1 log k = lim sup log k = log 2 ω(n) fk (n) fk (n) n→∞ Ω(n)

(17)

lim inf n→∞

Theorem 5. lim inf n→∞

σk∗ (p) = 2 for a prime p, and use inequality (10). fk (p) For the function D∗ (n) = 2Ω(n) remark that Proof. Remark that

lim sup log(f (n)D∗ (n)/n) = +∞.

(18)

n→∞

Indeed, by log(f (n)D∗ (n)) =

r X

ai log(pi +1) and by the known inequality

i=1

log(pi + 1) > log pi + 2/(2pi + 1) it follows ∗

log(f (n)D (n)/n) >

r X i=1

r

1X 2ai /(pi + 1) > 1/pi . 2

385

i=1

(19)

r X i=1

If one selects for pi , i = 1, r, the sequence of consecutive primes, then as 1/pi → ∞ as r → ∞, by (19) we get the weaker result (18). Theorem 6. There exists a sequence (nk ) such that log f (nk )D∗ (nk )/nk À log nk .

(20)

Proof. Let nk = 2k , in which case we have µ ¶k µ ¶k µ ¶log nk / log 2 3 1 3 k f (nk )D (nk )/nk = 2 = , 2 2 2 ∗

hence log f (nk )D∗ (nk )/nk = (log nk )

log 3/2 À log nk . log 2

Inequality (19) suggests the introduction of the arithmetic function h(n) =

n X ai i=1

pi

.

(21)

Theorem 7. For all n ≥ 2 one has 4 h(n) < log f (n)D∗ (n)/n < h(n). 5

(22)

p Proof. By the known inequality log(x + 1) < log x + 1/ x(x + 1), we get log f (n)D∗ (n)/n
0, m, k ≥ 0

(27)

where the inequality is strict, when at least one of k and m is ≥ 1. Apply (27) to x = pi , in order to obtain fk+n (n) ≥ fk (n)fm (n),

k, m ≥ 1,

which shows that (fk (n))k is supermultiplicative. The algebraic inequality xk+m + 1 ≤ xm , xk + 1

x > 0, m, k ≥ 0

(28)

and relation (1) give at once that fk+m (n) ≤ nm fk (n),

k, m ≥ 1.

(29)

For m = 1 inequality (29) gives fk+1 (n)/nk+1 ≤ fk (n)/nk , so assertion (26) is proved, too. 6. Finally, we consider certain divisibility problems. An important property for multiplicative functions, which is valid for f (n), too, is the following Theorem 9. If m|n (m divides n), then f (m) ≤ f (n). (30) Remark. The values of f (n) are not integers for all n. For example, f (2) = 3 . If n is odd, then all prime factors of n are odd, thus n is an integer. Let 2 n = 29 m, with (m, 2) = 1. Then by multiplicative property of f one has µ ¶9 3 f (n) = f (m), so 2a must divide f (m). Let p1 , . . . , pt , 1 ≤ t ≤ r, be all 2 prime factors of n which are of the form pi = 2mi +1 Mi − 1 (Mi odd, mi ≥ 0, 1 ≤ i ≤ t). Theorem 10. If m =

r Y

pbi i , pi odd, then the necessary and sufficient

i=1

condition for f (n) to be an integer is that r X

bi mi ≥ a,

i=1

where 2a kn. 387

(31)

References [1] T. Apostol, Introduction to analytic number theory, Springer Verlag, 1976. [2] E. Cohen, Arithmetical functions associated with the unitary divisors of an integer, Math. Z., 34(1960), 66-80. [3] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, Oxford Univ. Press, 1960. [4] M.J. deKoninck, On a class of arithmetical functions, Duke Math. J., 39(1972), 807-818. [5] A. Makowski, A. Schinzel, On the functions σ(n) and ϕ(n), Colloq. Math., 8(1964), 95-99. [6] J. Morgado, On the arithmetical function σk∗ , Portugal Math., 23(1964), 35-40. [7] C. Pomerance, On the composition of the arithmetic functions σ and ϕ, Colloq. Math., 58(1989), 11-15. [8] J. Sándor, Remarks on the functions σ(n) and ϕ(n), Babe¸s-Bolyai Univ. Preprint Nr.7, 1989, 7-12. [9] J. Sándor, L. Tóth, On certain number-theoretic inequalities, Fib. Quart. 28(1990), 255-258. [10] J. Sándor, An application of the Jensen-Hadamard inequality, Nieuw Arch. Wiskunde, 8(1990), no.4, 63-66. [11] J. Sándor, On the arithmetical functions dk (n) and d∗k (n), Portugal Math., 53(1996), no.1, 107-115. [12] J. Sándor, Handbook of number theory, II, Springer Verlag, 2005.

11

A generalization of Ruzsa’s theorem

Recently [1] we have published a new proof of a theorem of Ruzsa, which states that all even numbers a and b satisfying σ(a) = 2b, σ(b) = 2a (the so-called ”lovely pairs”) are given by a = 2k (2q+1 − 1),

b = 2q (2k+1 − 1),

388

where 2k+1 − 1 and 2q+1 − 1 are both prime numbers. The aim of this note is to show, that the methods of [1] enable us to give the following generalization. Theorem. Let f : N∗ → N∗ be a given multiplicative arithmetic function which satisfies the following properties: 1) f (2k ) is odd; 2) f (mn) ≥ mf (n) for all m, n ≥ 1, with equality only for m = 1; 3) f [f (2k )] ≥ 2k+1 for all k ≥ 1, with equality only for k ∈ A where A ⊂ N∗ . Then all even solutions of the system ½ f (a) = 2b f (b) = 2a in positive integers are given by a = 2k f (2a ), b = 2q f (2k ), where k, q ∈ A. If there is strict inequality in at least one of 2) and 3), then the system doesn’t have even solutions. Proof. Let a = 2k A, b = 2q B, where A and B are odd numbers. Since f is multiplicative, f (a) = f (2k )f (A), f (b) = f (2k )f (B), so f (2k )f (A) = 2q+1 B, f (2q )f (B) = 2k+1 A. By 1) f (2k ) is odd, so f (2k ) must divide B, i.e. B = f (2k )B 0 . Similarly, A = f (2q )A0 , where A0 , B 0 are odd integers. These give f (A) = 2q+1 B 0 , f (B) = 2k+1 A0 . By using relations 2) and 3) one can deduce: 2q+1 B = f (2k )f (A) = f (2k )f [A0 f (2q )] ≥ f (2k )A0 f [f (2q )] ≥ f (2k )A0 2q+1 . Since B = f (2k )B 0 , this gives B 0 ≥ A0 . In a completely analogous way one can obtain A0 ≥ B 0 . Thus A0 = B 0 , so A = f (2q )S, B = f (2k )S. Now, f (A) = 2q+1 S, f (B) = 2k+1 S give f [f (2q S)] = 2q+1 S, f [f (2k )S] = 2k+1 S. But f [f (2q )S] ≥ Sf [f (2q )] ≥ S · 2q+1 . Therefore, we must have equality, so S = 1 and q ∈ A. Similarly, k ∈ A, and the theorem is proved. When f (n) = σ(n), Ruzsa’s theorem is reobtained (select A = set of primes). When f (n) = nσ(n), all conditions are satisfied, with strict inequality in 3), so in this case the system is not solvable.

References [1] J. Sándor, On Ruzsa’s lovely pairs, Octogon Math. Mag., 12(2004), no.1, 287-289.

12

On the monotonicity of the sequence (σk /σk∗ )

1. Introduction Let n > 1 be a positive integer and k ≥ 0 a nonnegative integer. A divisor d of n is called a unitary divisor of n, if (d, n/d) = 1. Let σ ∗ (n) be the sum of 389

unitary divisors of n, i.e. X

σ ∗ (n) =

d.

(1)

r Y σ (n) = (pai i + 1),

(2)

d|n,(d,n/d)=1

Then it is well-known (see e.g. [1], [8]) that ∗

i=1

where n =

n Y

pai i is the prime factorization of n > 1 (pi distinct primes, ai ≥ 1

i=1

positive integers). More generally, if σk∗ (n) is the sum of kth powers of unitary divisors of n (i.e. (1) generalized to dk in place of d in the sum), then, similarly to (2), one has r Y ∗ (3) σk (n) = (pikai + 1). i=1

We note that, for k = 0 we get the number d∗ (n) = σ0∗ (n) of unitary divisors of n, when (3) gives d∗ (n) = 2r = 2ω(n) ,

(4)

where ω(n) = r denotes the number of unitary divisors of n. The similar formulae for the (classical) sum of divisors of n are the well-known (see e.g. [2], [9], [7]) r Y σ(n) = (pai i +1 − 1)/(pi − 1), (5) i=1

resp.

r Y k(a +1) σk (n) = (pi i − 1)/(pki − 1).

(6)

i=1

For k = 0, (6) provides the number d(n) of classical divisors of n: r Y (ai + 1). d(n) =

(7)

i=1

There are many results involving inequalities on these arithmetical functions. See e.g. [3]-[6]. For surveys of results, see e.g. [9], [8]. 390

2. Main results Langford ([9]) proved that µ σk (n) ≤ d(n)

nk + 1 2

¶ ,

(8)

while we proved ([10], [4], [5]) the stronger relation d(n)σk∗ (n) σk (n) ≤ ≤ d(n) 2ω(n)

µ

nk + 1 2

¶ .

(9)

The second inequality of (9) is a consequence of the elementary inequality r Y (xi + 1) ≤ 2r−1 i=1

Ã r Y

! xi + 1

(xi ≥ 1, r ≥ 1)

(10)

i=1

i applied to xi = pka i , r = ω(n), and using relation (3). Remark that the first inequality of (9) may be written also as

σ0 (n) σk (n) ≤ ∗ . ∗ σk (n) σ0 (n)

(11)

Our aim is to give a generalization of (11) as follows: µ ¶ σk (n) Theorem. For all fixed n ≥ 1, the sequence is monotone σk∗ (n) k≥0 decreasing. Proof. We have to prove that σk (n) σl (n) ≤ ∗ for all k ≥ l ≥ 0. ∗ σk (n) σl (n)

(12)

By (3) and (6), fk (n) =

σk (n)/σk∗ (n)

r Y k(a +1) i (pi i − 1)/(pki − 1)(pka + 1), = i i=1

so to prove that fk (n) ≤ fl (n) for k ≥ l, it will be sufficient to show that pl(a+1) − 1 pk(a+1) − 1 ≤ , (pk − 1)(pka + 1) (pl − 1)(pla + 1) 391

k ≥ l ≥ 0, p ≥ 2.

(13)

Put pk = x, pl = y, where x > y ≥ 1. After some elementary transformations (which we omit here) it can be shown that (13) becomes equivalent to xa − y a (xy)a − 1 ≤ (x > y ≥ 1). (14) x−y xy − 1 For y = 1, relation (14) is trivial, so we may suppose y ≥ 2. Now, remark that xa − y a = xa−1 + xa−2 y + · · · + xy a−2 + y a−1 ≤ axa−1 , x−y by y < x and a ≥ 1. On the other hand, we will prove that (xy)a − 1 ≥ axa−1 . xy − 1

(15)

This is equivalent to (xy)a − 1 ≥ axa y − axa−1 , or xa y(y a−1 − a) + axa−1 − 1 ≥ 0. Here axa−1 − 1 ≥ a − 1 ≥ 0, and y a−1 − a ≥ 2a−1 − a ≥ 0 for all a ≥ 1, so the result follows. By (15), and the above remark, inequality (14) is established. By (13), the inequality (12) follows, so the theorem is proved. Remarks. 1) For l = 0, k ≥ 0 arbitrary, we reobtain relation (11). 2) For l = 1, k ≥ 1 we get σ(n) σk (n) ≤ ∗ , ∗ σk (n) σ (n)

(16)

which offers an improvement of (11) for k ≥ 1, since by (11) applied to k = 1, and by (16), one has σ(n) σ0 (n) d(n) σk (n) ≤ ∗ ≤ ∗ = ∗ . ∗ σk (n) σ (n) σ0 (n) d (n)

(17)

For other improvements of the right side of (17), see [5].

References [1] E. Cohen, Arithmetical functions associated with the unitary divisors of an integer, Math. Z. 74(1960), 66-80. 392

[2] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, Oxford Univ. Press, 1960. [3] J. Sándor, An application of the Jensen-Hadamard inequality, Nieuw Arch. Wiskunde, Serie 4, 8(1990), no. 1, 43-66. [4] J. Sándor, On an inequality of Klamkin with arithmetical applications, Int. J. Math. Ed. Sci. Technol. 25(1994), 157-158. [5] J. Sándor, On certain inequalities for arithmetic functions, Notes Numb. Th. Discr. Math. 1(1995), 27-32. [6] J. Sándor, On the arithmetical functions dk (n) and d∗k (n), Portugaliae Math. 53(1996), no. 1, 107-115. [7] J. Sándor, Geometric theorems, diophantine equations, and arithmetic functions, American Research Press, Rehoboth, New Mexico, 2002. [8] J. Sándor, Handbook of number theory II, Springer-Verlag, 2004. [9] J. Sándor, D. S. Mitrinović, Handbook of number theory, Kluwer Acad. Publ., 1996. [10] J. Sándor, L. Tóth, On certain number-theoretic inequalities, Fib. Quart. 28(1990), no. 3, 255-258.

13

A note on exponential divisors and related arithmetic functions

1. Introduction Let n > 1 be a positive integer, and n = pa11 . . . par r its prime factorization. A number d|n is called an exponential divisor (or e-divisor, for short) of n if d = pb11 . . . pbrr with bi |ai (i = 1, r). This notion has been introduced by E. G. Straus and M. V. Subbarao [1]. Let σe (n), resp. de (n) denote the sum, resp. number of e-divisors of n, and let σe (1) = de (1) = 1, by convention. A number n is called e-perfect, if σe (n) = 2n. For results and References involving e-perfect numbers, and the arithmetical functions σe (n) and de (n), see [4]. For example, it is well-known that de (n) is multiplicative, and de (n) = d(a1 ) . . . d(ar ), 393

(1)

where n = pa11 . . . par r is the canonical form of n, and d(a) denotes the number of (ordinary) divisors of a. The e-totient function ϕe (n), introduced and studied in [4] is multiplicative, and one has ϕe (n) = ϕ(a1 ) . . . ϕ(ar ), (2) where ϕ is the classical Euler totient function. Let σ(a) denote the sum of (ordinary) divisors of a. The product of edivisors of n, denoted by Te (n) has the following expression (see [9]): σ(a1 )d(a2 )...d(ar )

Te (n) = p1

r )d(a1 )...d(ar−1 ) . . . pσ(a r

(3)

A number n is called multiplicatively e-perfect if Te (n) = n2 . Based on (3), in [9] we have proved that n is multiplicatively e-perfect iff n can be written as n = pm , where σ(m) = 2m, and p is a prime. Two notions of exponentially-harmonic numbers have been recently introduced by the author in [11]. Finally, we note that for a given arithmetic function f : N∗ → N∗ , in [5], [6] we have introduced the minimum function of f by Ff (n) = min{k ≥ 1 : n|f (k)}

(4)

Various particular cases, including f (k) = ϕ(k), f (k) = σ(k), f (k) = d(k), f (k) = S(k) (Smarandache function), f (k) = T (k) (product of ordinary divisors), have been studied recently by the present author. He also studied the duals of these functions (when these have sense) defined by Ff∗ (n) = max{k ≥ 1 : f (k)|n}

(5)

See e.g. [10] and the References therein.

2. Main notions and results The aim of this note is to introduce certain new arithmetic functions, related to the above considered notions. Since for the product of ordinary divisors of n one can write T (n) = nd(n)/2 ,

(6)

trying to obtain a similar expression for Te (n) of the product of e-divisors of n, by (3) the following can be written: Theorem 1. Te (n) = (t(n))de (n)/2 , (7) 394

where de (n) is the number of exponential divisors of n, given by (1); while the arithmetical function t(n) is given by t(1) = 1 σ(a )

2 d(a 1)

t(n) = p1

1

2

σ(ar )

. . . pr d(ar )

(8)

n = pa11 . . . par r being the prime factorization of n > 1. Proof. This follows easily by relation (3), and the definition of t(n) given by (8). Remark. For multiplicatively perfect numbers given by T (n) = n2 , see [7]. For multiplicatively deficient numbers, see [8]. Remark that de (n) ≤ d(n) (9) for all n, with equality only for n = 1. Indeed, by d(a) < a + 1 for a ≥ 2, via (1) this is trivial. On the other hand, the inequality t(n) ≤ n

(10)

is not generally valid. Let e.g. n = pq11 . . . pqrr , where all qi (i = 1, r) are primes. Then, by (8) t(n) = pq11 +1 . . . pqrr +1 = (p1 . . . pr )n > n. However, there is a particular case, when (10) is always true, namely suppose that ω(ai ) ≥ 2 for all i = 1, r (where ω(a) denotes the number of distinct prime factors of a). In σ(a) a [3] it is proved that if ω(a) ≥ 2, then < . This gives (10) with strict d(a) 2 inequality, if the above conditions are valid. Without any condition one can prove: Theorem 2. For all n ≥ 1 Te (n) ≤ T (n), with equality only for n = 1 and n = prime. Proof. The inequality to be proved becomes ³ ´d(a1 )...d(ar ) σ(a )/d(a1 ) r )/d(ar ) p1 1 . . . pσ(a ≤ (pa11 . . . par r )(a1 +1)...(ar +1)/2 r

(11)

(12)

We will prove that σ(a1 ) a1 (a1 + 1) . . . (ar + 1) d(a1 ) . . . d(ar ) ≤ d(a1 ) 2 with equality only if r = 1 and a1 = 1. Indeed, it is known that (see [2]) a1 + 1 σ(a1 ) ≤ , with equality only for a1 = 1 and a1 = prime. On the d(a1 ) 2 395

other hand, d(a1 ) . . . d(ar ) ≤ a1 (a2 + 1) . . . (ar + 1) is trivial by d(a1 ) ≤ a1 , d(a2 ) < a2 + 1, . . . , d(ar ) < ar + 1, with equality only for a1 = 1 and r = 1. Thus (12) follows, with equality for r = 1, a1 = 1, so n = p1 = prime for n > 1. Remark. In [4] it is proved that ϕe (n)de (n) ≥ a1 . . . ar

(13)

a1 ar ... ≥ 2r if all ai (i = 1, r) are even, since ϕ(a1 ) ϕ(ar ) a it is well-known that ϕ(a) ≤ for a = even. Since d(n) = (a1 +1) . . . (ar +1) ≤ 2 2a1 . . . 2ar = 2a1 +···+ar = 2Ω(n) (where Ω(n) denotes the total number of prime divisors of n), by (9) one can write: Now, by (2), de (n) ≥

2ω(n) ≤ de (n) ≤ 2Ω(n) ,

(14)

if all ai are even, i.e. when n is a perfect square (right side always). Similarly, in [4] it is proved that ϕe (n)de (n) ≥ σ(a1 ) . . . σ(ar )

(15)

when all ai (i = 1, r) are odd. Let all ai ≥ 3 be odd. Then, since σ(ai ) ≥ ai + 1 (with equality only if ai = prime), (15) implies ϕe (n)de (n) ≥ d(n),

(16)

which is a converse to inequality (9). √ √ 2 a 2 a Let now introduce the arithmetical function t1 (n) = p1 1 . . . pr r , t1 (1) = 1 and let γ(n) = p1 . . . pr denote the ”core” of n (see [2]). Then: Theorem 3. t1 (n) ≤ t(n) ≤ nγ(n) for all n ≥ 1.

(17)

Proof. This follows at once by the known double-inequality √ a+1 σ(a) ≤ , a≤ d(a) 2

(18)

with equality for a = 1 on the left side, and for a = 1 and a = prime on the right side. Therefore, in (17) one has equality when n is squarefree, while on the right side if n is squarefree, or n = pq11 . . . pqrr with all qi (i = 1, r) primes. Clearly, the functions t1 (n), t(n) and γ(n) are all multiplicative. 396

Finally, we introduce the minimum exponential totient function by (4) for f (k) = ϕe (k): Ee (n) = min{k ≥ 1 : n|ϕe (k)}, (19) where ϕe (k) is the e-totient function given by (2). Let E(n) = min{k ≥ 1 : n|ϕ(k)} be the Euler minimum function (see [10]). The following result is true: Theorem 4. Ee (n) = 2E(n) for n > 1.

(20)

(21)

Proof. Let k = pα1 1 . . . pαs s . Then k ≥ 2α1 +···+αs ≥ 2s . Let s be the least integer with n|ϕ(s) (i.e. s = E(n) by (20)). Clearly ϕe (2s ) = ϕ(s), so k = 2s is the least k ≥ 1 with property n|ϕe (k). This finishes the proof of (21). For properties of E(n), see [10]. Remark. It is interesting to note that the ”maximum e-totient”, i.e. Ee∗ (n) = max{k ≥ 1 : ϕe (k)|n}

(22)

is not well defined. Indeed, e.g. for all primes p one has ϕe (p) = 1|n, and Ee∗ (p) = +∞, so Ee∗ (n) given by (22) is not an arithmetic function.

References [1] E. G. Straus and M. V. Subbarao, On exponential divisors, Duke Math. J. 41(1974), 465-471. [2] D. S. Mitronović and J. Sándor, Handbook of number theory, Kluwer Acad. Publ., 1995. [3] J. Sándor, On the Jensen-Hadamard inequality, Nieuw Arch Wiskunde (4)8(1990), 63-66. [4] J. Sándor, On an exponential totient function, Studia Univ. Babe¸sBolyai, Math. 41(1996), no. 3, 91-94. [5] J. Sándor, On certain generalizations of the Smarandache functions, Notes Number Th. Discr. Math. 5(1999), no. 2, 41-51. [6] J. Sándor, On certain generalizations of the Smarandache function, Smarandache Notion J. 11(2000), no. 1-3, 202-212. 397

[7] J. Sándor, On multiplicatively perfect numbers, J. Ineq. Pure Appl. Math. 2(2001), no. 1, Article 3, 6 pp. (electronic). [8] J. Sándor, Geometric theorems, diophantine equations, and arithmetic functions, American Research Press, Rehoboth, 2002. [9] J. Sándor, On multiplicatively e-perfect numbers, to appear. [10] J. Sándor, On the Euler minimum and maximum functions, to appear. [11] J. Sándor, On exponentially harmonic numbers, to appear.

14

The Euler minimum and maximum functions 1. The Euler minimum function is defined by E(n) = min{k ≥ 1 : n|ϕ(k)}

(1)

It was introduced by P. Moree and H. Roskam [5]; and independently by J. Sándor [7], as a particular case of the more general function FfA (n) = min{k ∈ A : n|f (k)} (A ⊂ N∗ ),

(2)

where f : N∗ → N∗ is a given function, and A is a given set of positive integers. For A = N∗ , f = ϕ (Euler’s totient), one obtains the function E given by (1) (denoted also as Fϕ in [7]). Since by Dirichlet’s theorem on arithmetical progression, there exists a ≥ 1 such that k = an + 1 = prime, by ϕ(k) = an˙:n, so E(n) is well defined. We note that for A = N∗ , f (k) = k! one reobtains the Smarandache function S(n) = min{k ≥ 1 : n|k!}, (3) while for A = P = {2, 3, 5, . . . } = set of all primes, f (k) = k!, (2) gives a new function, denoted by us as P (n): P (n) = min{k ∈ P : n|k!}

(4)

We note that this function should not be confused with the greatest prime divisor of n (denoted also sometimes by P (n)). For properties of this function, see [6], [7]. There is a dual of (2) (see [7]), namely GA g (n) = max{k ∈ A : g(k)|n}, 398

(5)

where g : N∗ → N∗ , A ⊂ N∗ are given, if this is well defined. For A = N∗ , g(k) = k!, this has been denoted by us by S∗ (n), and called as the dual of the Smarandache function: S∗ (n) = max{k ≥ 1 : k!|n}

(6)

For properties of this function, see [6]. See also F. Luca [4], where a conjecture of the author has been proved, and M. Le [3] for a recent new proof. See also K. Atanassov [1]. For A = N∗ , g(k) = ϕ(k) one obtains the dual E∗ (n) of the Euler minimum function, which we shall call as the Euler maximum function: E∗ (n) = max{k ≥ 1 : ϕ(k)|n} (7) √ Since for k > 6, ϕ(k) > k, clearly k < n2 , so E∗ (n) ≤ n2 < ∞. Generally, for A = N∗ , let us write simply FfA (n) = Ff (n), GA g (n) = Gf (n). 2. First we prove the following property of the Euler minimum function: Theorem 1. If pi (i = 1, r) are distinct primes, and αi ≥ 1 are integers, then ! Ã r Y α αi (8) max{E(pi ) : i = 1, r} ≤ E pi i ≤ [E(pα1 1 ), . . . , E(pαr r )], i=1

where [, . . . , ] denotes l.c.m. Proof. For simplicity we shall prove (8) for r = 2. Let pα , q β be distinct prime powers. Then E(pα q β ) = min{k ≥ 1 : pα q β |ϕ(k)} = k0 , so pα q β |ϕ(k0 ), which is equivalent to pα |ϕ(k0 ), q β |ϕ(k0 ), thus k0 ≥ E(pα ), k0 ≥ E(q β ), imβ ) ≥ max{E(pα ), E(q β )}. It is immediate that the same proof plying E(pα q³ Y ´ applies to E pα ≥ max{E(pα )}, where pα are distinct prime powers. Therefore, the left side of (8) follows. Now, let E(pα ) = k1 , E(q β ) = k2 , implying pα |ϕ(k1 ), q β |ϕ(k2 ). Let [k1 , k2 ] = g. Since k1 |g, one has ϕ(k1 )|ϕ(g) (by a known property of the function ϕ). Similarly, since k2 |g, one can write ϕ(k2 )|ϕ(g). Thus pα |ϕ(k1 )|ϕ(g) and q β |ϕ(k2 )|ϕ(g), yielding pα q β |ϕ(g). By the definition (1) this gives g ≥ E(pα q β ), i.e. [E(pα ), E(q β )] ≥ E(pα q β ), so the right side of (8) for r = 2 is proved. The general case follows exactly the same lines. Remark 1. The above proof shows that the left side of (8) holds true for any function f (for which Ff is well defined), so we get Ã r ! Y max{Ff (pαi i ) : i = 1, r} ≤ Ff pαi i (9) i=1

399

For the right side of (8), with the same proof the following is valid: if f has the property a|b ⇒ f (a)|f (b) (a, b ≥ 1), (10) then Ff

Ã r Y

! pαi i

≤ [Ff (pα1 1 ), . . . , Ff (pαr r )]

(11)

i=1

Now, if one replaces (10) with a stronger property, then a better result will be true: Theorem 2. Assume that f satisfies the following property a ≤ b ⇒ f (a)|f (b) Then Ff

Ã r Y

(a, b ≥ 1)

(12)

! pαi i

= max{Ff (pαi i ) : i = 1, r}

(13)

i=1

Proof. By taking into account of (9), one needs only to show that the reverse inequality is true. For simplicity, let us take again r = 2. Let Ff (pα ) = m, Ff (q β ) = n, with m ≤ n. Then the definition (2) of Ff implies that pα |f (m), q β |f (n). By (12) one has f (m)|f (n), so pα |f (m)|f (n). We have pα |f (n), q β |f (n), so pα q β |f (n). But this implies n ≥ Ff (pα q β ), i.e. max{Ff (pα ), Ff (q β )} ≥ Ff (pα q β ). The general case follows exactly the same lines. s r Y Y β qj j , (pi , qj ) = 1, Remark 2. If (a, b) = 1, by writing a = pαi i , b = i=1

j=1

it follows that β

Ff (ab) = max{E(pαi i ), E(qj j ) : i = 1, r, j = 1, s} = β

= max{max{E(pαi i ) : i = 1, r}, max{E(qj j ) : j = 1, s}} = = max{Ff (a), Ff (b)}, so: Ff (ab) = max{Ff (a), Ff (b)} for (a, b) = 1

(14)

When f (n) = n!, then clearly (12) is true, so (14) gives: S(ab) = max{S(a), S(b)} for (a, b) = 1, discovered by F. Smarandache [9]. 400

(15)

3. The Euler minimum function must be studied essentially (by Theorem 1) for prime powers pα . For values of E(p), E(p2 ), etc., see [5]. On the other hand, for each prime p ≥ 3 one has E(p − 1) = p

(16)

Indeed, if (p − 1)|ϕ(k), then p − 1 ≤ ϕ(k). Since ϕ(k) ≤ k − 1 for k ≥ 3, one has p ≤ k. Now k = p gives ϕ(p) = p − 1, giving (16). The values of the Euler maximum function E∗ given by (7) however are even difficult to calculate in some cases. This function doesn’t seem to have been studied up to now. Clearly E∗ (1) = 2 since ϕ(1) = 1, ϕ(2) = 1. One has E∗ (2) = 6 since ϕ(6) = 2, 2|2, and it is well-known that ϕ(n) ≥ 4 for n ≥ 7. Now let p ≥ 3 be a prime. Since ϕ(k)|p implies ϕ(k) = 1 or ϕ(k) = p, for p ≥ 3 the last equality is impossible for ϕ(k) is even for all k ≥ 3, we can have only ϕ(k) = 1 and kmax = 2. Actually since for k ≥ 3, ϕ(k) is even, ϕ(k)|n is impossible for n = odd, so remains k ≤ 2, and kmax = 2. We have½proved: 6, if p = 2 Theorem 3. One has E∗ (1) = 2, E∗ (p) = for all primes 2, if p ≥ 3 p; and E∗ (n) = 2 for all n ≥ 1 odd. For all n ≥ 2 one has E∗ (n) ≥ 2. (17) The last inequality is a consequence of ϕ(2) = 1 and the definition (7). The value 2 is taken infinitely often, but the same is true for the value 6: Theorem 4. For all α ≥ 1 one has E∗ (2 · 7α ) = 6

(18)

Proof. If ϕ(k)|(2 · 7α ), then assuming k ≥ 3, as ϕ(k) is even, we can only have ϕ(k) = 2 or ϕ(k) = 2 · 7a where 1 ≤ a ≤ α. Now, A. Schinzel [8] has shown that the equation ϕ(x) = 2 · 7a is not solvable for any a ≥ 1. Thus, it remains ϕ(k) = 2 and the maximal value of k ≥ 3 is k = 6. This finishes the proof of (18). Remark. One has similarly E∗ (2 · 52α ) = 6 for any α ≥ 1. (19) The function E∗ can take greater values, too; the values at powers of 2 is shown by the following theorem: Theorem 5. E∗ (2m ) = k, where k is the greatest number which can be α αr written as k = 2α p1 . . . pr , with p1 = 22 1 + 1, . . . , pr = 22 + 1 distinct Fermat primes, and where α = a + 1 − (2k1 + · · · + 2kr ), with k1 , . . . , kr ≥ 0, 0 ≤ a ≤ m. (20) m a Proof. Since ϕ(k)|2 , clearly ϕ(k) = 2 , where 0 ≤ a ≤ m. Now let k = 2α pα1 1 . . . pαr r with p1 , . . . , pr distinct odd primes. Since ϕ(k) = 401

2α−1 pα1 1 −1 . . . pαr r −1 (p1 − 1) . . . (pr − 1) = 2a , we must have α1 − 1 = · · · = αr − 1 = 0 and p1 − 1 = 2a1 , . . . , pr − 1 = 2ar with α − 1 + a1 + · · · + ar = a. This gives p1 = 2a1 + 1, . . . , pr = 2ar + 1. Since p1 is prime, it is well-known that it is a Fermat prime, so a1 = 2k1 , etc., and the theorem follows. Remark 3. For m = 2 we get α ≤ 3 − (2k1 + · · · + 2kr ), so with k1 = 0 (when p1 = 3), we get k = 22 · 3 = 12. Another value would be k = 2 · 5 = 10, so we get E∗ (4) = 12 Similarly, for m = 3, E∗ (8) = 30 If can be shown also that E∗ (16) = 60,

E∗ (32) = 120,

E∗ (64) = 240,

E∗ (128) = 510, etc.

However, since the structure (or the cardinality) of the Fermat primes is not well-known, there are problems also with the calculation of great values of E∗ (2m ). The function E∗ (n) can take arbitrarily large values, since one has: Theorem 6. For all m ≥ 1 the following inequality is true: E∗ (m!) ≥

(m!)2 ϕ(m!)

(21)

Proof. It is known (see e.g. [2]) that the equation ϕ(x) = m!

(22)

admits the solution x = (m!)2 /ϕ(m!). Now, since ϕ(x) = m!|m!, clearly E∗ (m!) ≥ x, giving inequality (21). E∗ (m!) = +∞ (23) Corollary. lim m→∞ m! m! → ∞ as m → ∞. Proof. Indeed, it is well-known (see e.g. [10]) that ϕ(m!) By (21), this implies (23).

References [1] K. T. Atanassov, Remark on J´ ozsef S´ andor and Florian Luca’s theorem, C. R. Acad. Bulg. Sci. 55(2002), no. 10, 9-14. [2] P. Erdös, Amer. Math. Monthly 58(1951), p. 98. 402

[3] M. Le, A conjecture concerning the Smarandache dual function, Smarandache Notion J. 14(2004), 153-155. [4] F. Luca, On a divisibility property involving factorials, C. R. Acad. Bulg. Sci. 53(2000), no. 6, 35-38. [5] P. Moree, H. Roskam, On an arithmetical function related to Euler’s totient and the discriminantor, Fib. Quart. 33(1995), 332-340. [6] J. Sándor, On certain generalizations of the Smarandache function, Smarandache Notions J. 11(2000), no. 1-3, 202-212. [7] J. Sándor, On certain generalizations of the Smarandache function, Notes Number Theory Discr. Math. 5(1999), no. 2, 41-51. [8] A. Schinzel, Sur l’équation ϕ(x) = m, Elem. Math. 11(1956), 75-78. [9] F. Smarandache, A function in the number theory, An. Univ. Timi¸soara, Ser. Mat., 38(1980), 79-88. [10] J. Sándor, On values of arithmetical functions at factorials, I, Smarandache Notions J., 10(1999), 87-94.

15

The Smarandache minimum and maximum functions

1. Let f : N∗ → N be a given arithmetic function and A ⊂ N a given set. The arithmetical function FfA (n) = min{k ∈ A : n|f (k)}

(1)

has been introduced in [4] and [5]. For A = N, f (k) = k! one obtains the Smarandache function; for A = N∗ , A = P = {2, 3, 5, . . . } = set of all primes, one obtains a function P (n) = min{k ∈ P : n|k!}

(2)

For properties of this function, see [4], [5]. The ”dual” function of (1) has been defined by GA g (n) = max{k ∈ A : g(k)|n}, 403

(3)

where g : N∗ → N is a given function, and A ⊂ N is a given set. Particularly, for A = N∗ , g(k) = k! one obtains the dual of the Smarandache function, S∗ (n) = max{k ≥ 1 : k!|n}

(4)

For properties of this function, see [4], [5]. F. Luca [3], K. Atanassov [1] and L. Le [2] have proved in the affirmative a conjecture of the author. For A = N∗ and f (k) = g(k) = ϕ(k) in (1), resp. (3) one obtains the Euler minimum, resp. maximum-functions, defined by E(n) = min{k ≥ 1 : n|ϕ(k)},

(5)

E∗ (n) = max{k ≥ 1 : ϕ(k)|n}

(6)

For properties of these functions, see [6]. When A = N∗ , f (k) = d(k) = number of divisors of k, one obtains the divisor minimum function (see [4], [5], [7]) D(n) = min{k ≥ 1 : n|d(k)}

(7)

It is interesting to note that the divisor maximum function (i.e. the ”dual” of D(n)) given by D∗ (n) = max{k ≥ 1 : d(k)|n} (8) is not well-defined! Indeed, for any prime p one has d(pn−1 ) = n|n and pn−1 is unbounded as p → ∞. For a finite set A, however D∗A (n) does exist. On the other hand, it has been shown in [4], [5] that Σ(n) = min{k ≥ 1 : n|σ(k)}

(9)

(denoted there by Fσ (n)) is well defined. (Here σ(k) denotes the sum of all divisors of k). The dual of the sum-of-divisors minimum function is Σ∗ (n) = max{k ≥ 1 : σ(k)|n}

(10)

Since σ(1) = 1|n and σ(k) ≥ k, clearly Σ∗ (n) ≤ n, so this function is well-defined (see [8]). 2. The Smarandache minimum function will be defined for A = N∗ , f (k) = S(k) in (1). Let us denote this function by Smin : Smin (n) = min{k ≥ 1 : n|S(k)}

(11)

Let us assume that S(1) = 1, i.e. S(n) is defined by (1) for A = N∗ , f (k) = k!: S(n) = min{k ≥ 1 : n|k!} (12) 404

Otherwise (i.e. when S(1) = 0) by n|0 for all n, by (11) one gets the trivial function Smin (n) ≡ 0. By this assumption, however, one obtains a very interesting (and difficult) function Smin given by (11). Since n|S(n!) = n, this function is correctly defined. The Smarandache maximum function will be defined as the dual of Smin : Smax (n) = max{k ≥ 1 : S(k)|n} (13) We prove that this is well-defined. Indeed, for a fixed n, there are a finite number of divisors of n, let i|n be one of them. The equation S(k) = i

(14)

is well-known to have a number of d(i!) − d((i − 1)!) solutions, i.e. in a finite number. This implies that for a given n there are at most finitely many k with S(k)|n, so the maximum in (13) is attained. Clearly Smin (1) = 1, Smin (2) = 2, Smin (3) = 3, Smin (4) = 4, Smin (5) = 5, Smin (6) = 9, Smin (7) = 7, Smin (8) = 32, Smin (9) = 27, Smin (10) = 25, Smin (11) = 11, etc., which can be determined from a table of Smarandache numbers: n 1 2 3 4 5 6 7 8 9 10 11 12 13 S(n) 1 2 3 4 5 3 7 4 6 5 11 4 13 n 14 15 16 17 18 19 20 21 22 23 24 25 S(n) 7 5 6 7 6 19 5 7 11 23 4 10 We first prove that: Theorem 1. Smin (n) ≥ n for all n ≥ 1, with equality only for n = 1, 4, p (p = prime). (15) Proof. Let n|S(k). If we would have k < n, then since S(k) ≤ k < n we would get S(k) < n, in contradiction with n|S(k). Thus k ≥ n, and taking minimum, the inequality follows. There is equality for n = 1 and n = 4. Let now n > 4. If n = p = prime, then p|S(p) = p, but for k < p, p - S(k). Indeed, by S(k) ≤ k < p this is impossible. Reciprocally, if min{k ≥ 1 : n|S(k)} = n, then n|S(n), and by S(n) ≤ n this is possible only when S(n) = n, i.e. when n = 1, 4, p (p = prime). Theorem 2. For all n ≥ 1, Smin (n) ≤ n! ≤ Smax (n).

405

(16)

Proof. Since S(n!) = n, definition (11) gives the left side of (16), while definition (13) gives the right side inequality. X 1 Corollary. The series is divergent, while the series Smin (n) n≥1 X 1 is convergent. Smax (n) n≥1 X X 1 1 Proof. Since ≤ = e − 1 by (16), this series is converSmax (n) n! n≥1 n≥1 X X X1 1 1 gent. On the other hand, ≥ = = +∞, so Smin (n) Smin (p) p p p prime

n≥1

the first series is divergent. Theorem 3. For all primes p one has Smax (p) = p!

(17)

Proof. Let S(k)|p. Then S(k) = 1 or S(k) = p. We prove that if S(k) = p, then k ≤ p!. Indeed, this follows from the definition (12), since S(k) = min{m ≥ 1 : k|m!} = p implies k|p!, so k ≤ p!. Therefore the greatest value of k is k = p!, when S(k) = p|p. This proves relation (17). Since S(p2 ) = 2p, for all primes p, from (11) and (13) one can deduce the first part of the following theorem: Theorem 4. For all primes p, Smin (2p) ≤ p2 ≤ Smax (2p),

(18)

and more generally; for all m ≤ p, Smin (mp) ≤ pm ≤ Smax (mp)

(19)

Proof. (19) follows by the known relation S(pm ) = mp if m ≤ p and the definitions (11), (13). Particularly, for m = 2, (19) reduces to (18). For m = p, (19) gives Smin (p2 ) ≤ pp ≤ Smax (p2 ) (20) The case when m is also an arbitrary prime is given in Theorem 5. For all odd primes p and q, p < q one has Smin (pq) ≤ q p ≤ pq ≤ Smax (pq) (21) holds also when p = 2 and q ≥ 5. 406

(21)

Proof. Since S(q p ) = pq and S(pq ) = qp for primes p and q, the extreme inequalities of (21) follow from the definitions (11) and (13). For the inequality ln x q p < pq remark that this is equivalent to f (p) > f (q), where f (x) = x 1 − ln x 0 (x ≥ 3). Since f (x) = = 0 ⇔ x = e immediately follows that f is x2 strictly decreasing for x ≥ e = 2.71 . . . From the graph of this function, since ln 2 ln 4 ln 2 ln 3 ln 2 ln q = we get that < , but > for q ≥ 5. Therefore (21) 2 4 2 3 2 q holds also when p = 2 and q ≥ 5. Indeed, f (q) ≤ f (5) < f (4) = f (2). Remark. For all primes p, q Smin (pq) ≤ min{pq , q p } and Smax (pq) ≥ max{pq , q p }.

(22)

For p = q this implies relation (21). Proof. Since S(q p ) = S(pq ) = pq, one has Smin (pq) ≤ pq ,

Smin (pq) ≤ q p ,

Smax (pq) ≥ pq ,

Smax (pq) ≥ q p .

References [1] K. Atanassov, Remark on J´ ozsef S´ andor and Florian Luca’s theorem, C. R. Acad. Bulg. Sci. 55(2002), no. 10, 9-14. [2] M. Le, A conjecture concerning the Smarandache dual function, Smarandache Notions J. 14(2004), 153-155. [3] F. Luca, On a divisibility property involving factorials, C. R. Acad. Bulg. Sci. 53(2000), no. 6, 35-38. [4] J. Sándor, On certain generalizations of the Smarandache function, Notes Number Theory Discr. Math. 5(1999), no. 2, 41-51. [5] J. Sándor, On certain generalizations of the Smarandache function, Smarandache Notions J., 11(2000), no. 1-3, 202-212. [6] J. Sándor, On the Euler minimum and maximum functions, RGMIA Research Report Collection 8(1), Article 1, 2005. [7] J. Sándor, A note on the divisor minimum function, Octogon Math. Mag., 12(2004), no.2A, 273-275. [8] J. Sándor, The sum-of-divisors minimum and maximum functions, RGMIA Research Report Collection, 8(1), Article 20, 2005. 407

16

The pseudo-Smarandache minimum and maximum functions

1. Introduction K. Kashihara [2] defined the pseudo-Smarandache function Z by ½ ¾ m(m + 1) Z(n) = min m ≥ 1 : n| 2

(1)

For properties of this function see e.g. [2], [1], [6]. The dual of the pseudoSmarandache function has been introduced and studied by the author (see [5], [6): ½ ¾ m(m + 1) Z∗ (n) = max m ≥ 1 : |n (2) 2 More generally, for functions f, g : N∗ → N∗ , the author [3], [4] (see also [6]) has defined the functions Ff (n) = min{k ≥ 1 : n|f (k)},

(3)

Gg (n) = max{k ≥ 1 : g(k)|n},

(4)

if these functions are well-defined. Various particular cases, including f (k) = g(k) = ϕ(k), f (k) = g(k) = σ(k), f (k) = d(k), f (k) = S(k) (Smarandache function), f (k) = T (k) (product of ordinary divisors) have been studied by the author. See e.g. [7] and the References therein. The aim of this note is the initial study of the functions Z(n) and Z ∗ (n) given by (3), resp. (4) for the particular cases f (k) = g(k) = Z(k).

2. The pseudo-Smarandache minimum and maximum functions The pseudo-Smarandache minimum function will be defined by Z(n) = min{k ≥ 1 : n|Z(k)},

(5)

while the pseudo-Smarandache maximum function will be the dual of Z(n): Z ∗ (n) = max{k ≥ 1 : Z(k)|n} (6) The particular cases of (3) and (4) for f (k) = g(k) = Z∗ (k), given by (2) (i.e. the pseudo-Smarandache dual minimum and maximum functions) will be studied in another paper. 408

Recall the following known properties of Z(n) (see e.g. [6]) ½ 2n − 1, if n is even, Z(n) ≤ n − 1, if n is odd,

(7)

Z(2k ) = 2k+1 − 1,

(8)

Z(p) = p − 1 for p ≥ 3 prime,

(9)

Z(2p) = p − 1 for p ≡ 1 (mod 4) a prime, µ ¶ n(n + 1) Z = n for all n ≥ 1, 2 r 1 1 Z(n) ≥ − + 2n + for all n ≥ 1 2 4 Theorem 1. For all n ≥ 1, n+1 n(n + 1) ≤ Z(n) ≤ 2 2

(10) (11) (12)

(13)

µ

¶ n(n + 1) Proof. Since by (11), n|Z for all n, by (5) we get the right side 2 inequality of (13). On the other hand, remark that if n|Z(k), then n ≤ Z(k). n+1 . This implies From (7) we get Z(k) ≤ 2k − 1, so n ≤ 2k − 1, giving k ≥ 2 the left side inequality of (13). p n Corollary. lim Z(n) = 1. n→∞ Theorem 2. For all primes p, Z(p − 1) ≤ p

(14)

Z(p − 1) = p

(15)

If p ≡ 3 (mod 4), then

Proof. For p = 2, Z(1) = 1 < 2; let p ≥ 3. By (9) one has p − 1|Z(p) so (14) follows from definition (5). Now let p − 1|Z(k). If k is odd, then by p − 1 ≤ Z(k) ≤ k − 1 (see (7)) one gets p ≤ k. But Z(p) = p − 1. Further, if k < p is odd, then p ≤ k < p, which is impossible. If k is even, then Z(k) ≤ 2k − 1 and p − 1|Z(k) gives Z(k) = (p − 1)a ≤ 2k − 1 < 2p − 1. For a = 2 one can write 2(p − 1) = 2p − 2 ≤ 2k − 1 so 2k ≥ 2p − 1, 1 i.e. k ≥ p − , and k being integer, k ≥ p. Since p is odd, k ≥ p + 1, so for 2 409

k = p (in the odd case) we get a smaller value. For a = 1, Z(k) = p − 1. (n − 1)p We search for even k < p for which this is true. Since by (1), k| , we 2 get (p − 1)p = 2k · A (A = integer), so k|(p − 1)p. But p being prime, and k < p, clearly (k, p) = 1, so we must have k|(p − 1), i.e. p − 1 = km. Thus km(kn + 1) = 2kA, i.e. m(km + 1) = 2A. Since k is even, km + 1 is odd, so m must be even. From p − 1 = km it follows p ≡ 1 (mod 4). This contradicts the made assumption X on p. Corollary. 1/Z(p − 1) is divergent. Theorem 3. Z(2n − 1) = 2n−1 for all n ≥ 1 (16) Proof. The left side of (13) gives Z(2n − 1) ≥

2n − 1 + 1 = 2n−1 2

(∗)

On the other hand, by (8), Z(2n−1 ) = 2n − 1, so (2n − 1)|Z(2n−1 ), implying by (5) that Z(2n − 1) ≤ 2n−1 (∗∗) Relations (∗)X and (∗∗) imply (16). Corollary. 1/Z(2n − 1) is convergent. Theorem 4. n(n + 1) (17) 1 ≤ Z∗ (n) ≤ 2 Proof. n. Since by (12) this yields n ≥ r By (6), if Z(k)|n, thenµZ(k) ≤ ¶2 1 1 1 1 1 n(n + 1) − + 2k + , we get 2k + ≤ n + = n2 + n + , so k ≤ , 2 4 4 2 4 2 giving the right side of (17). The left side of (17) is obvious. p(p + 1) Theorem 5. Z∗ (p) = for all primes p. (18) 2 Proof. Let Z(k)|p. Since p is prime, we must have Z(k) = 1 (i.e. k = 1) or p(p + 1) is a solution. Z(k) = p. This equation is always solvable, since k = 2 p(p + 1) p(p + 1) So Z∗ (p) ≥ . On the other hand, (17) reads Z∗ (p) ≤ , so (18) 2 2 follows. Remark. The calculation of Z(p) seems much difficult. Theorem 6. If p ≡ 1 (mod 4) is a prime, then Z∗ (p − 1) ≥ 2p 410

(19)

Proof. By (10), Z(2p)|(p − 1), so (19) is a consequence of this fact and definition (6). Theorem 7. Z(Z∗ (n))|n|Z(Z(n)) for all n ≥ 1. (20) Proof. Let Z∗ (n) = k. Then by (6), Z(k)|n, so the left side of (20) follows. The right side follows similarly by (5). Corollary. Z(Z∗ (n)) ≤ n ≤ Z(Z(n)) for all n ≥ 1. (21)

References [1] C. Ashbacher, The pseudo-Smarandache function and the classical functions of Number Theory, Smarandache Notions J., 9(1998), No. 1-2, 7881. [2] K. Kashihara, Comments and Topics on Smarandache Notions and Problems, Erhus U.P., AZ, 1996. [3] J. Sándor, On certain generalizations of the Smarandache function, Notes Number Th. Discr. Math. 5(1999), no. 2, 41-51. [4] J. Sándor, On certain generalizations of the Smarandache function, Smarandache Notions J., 11(2000), no. 1-3, 202-212. [5] J. Sándor, On a dual of the pseudo-Smarandache function, Smarandache Notions J., 13(2002), no. 1-2-3, pp. 18-23. [6] J. Sándor, Geometric theorems, diophantine equations, and arithmetic functions, American Research Press, Rehoboth 2002. [7] J. Sándor, On the Euler minimum and maximum functions, RGMIA Research Report Collection, 8(1), Article 1, 2005.

411

412

Chapter 10

Miscellaneous themes ”... Say what you know, do what you must, come what may.” (Sofia Kovalevskaya)

”... The real purpose of mathematics is to be the means to illuminate reason and to exercise spiritual forces.” (August Crelle)

413

1

On a divisibility problem

1. The problem in the title is: Determine all positive integers x and y such that: xy−1 + y x−1 ≡ 0 (mod (x + y)). (1) In our opinion the most general solution of this divisibility cannot be never obtained, and only particular (special) cases are treatable. In what follows we shall indicate certain cases when (1) is soluble, or not. 2. First remark that (1) is always solvable when x = 2m, y = 2n with m > n; m + n = 2s , where s + 1 ≤ 2n − 1. Indeed, by substituting x = 2m, y = 2n in (1) we get xy−1 + y x−1 = (2m)2n−1 + (2n)2m−1 = 22n−1 (m2n−1 + 22m−2n n2m+1 ) which is divisible by 2(m + n) = 2 · 2s = 2s+1 where m > n and s + 1 ≤ 2n − 1. 3. Now, we will prove that (1) is not solvable if x = 2m + 1, y = 2n + 1 and m + n 6= 0 (mod 2). We shall use the known fact that the square of an odd number a is the form a2 ≡ 1 (mod 8). By this result xy−1 + y x−1 = (2m + 1)2n + (2n + 1)2m = (8M + 1) + (8N + 1) = 8K + 2 = 2(4K + 1). Since x + y = 2(m + n) + 2 = 2(m + n + 1); if m + n is odd, then m + n + 1 is even, which cannot divide an odd number (of the form 4K + 1). 4. Let x > y, y even and (x, y) = 1. Then (1) is true if and only if y x−y − 1 ≡ 0 (mod (x + y)).

(2)

Indeed, xy−1 + y x−1 = xy−1 + y y−1 + y x−1 − y y−1 = (xy−1 + y y−1 ) + y y−1 (y x−y − 1). Now, since y − 1 is odd, it is well-known that xy−1 + y y−1 is divisible by x + y. Since (xy−1 , x + y) = 1, by the above equality, (2) follows.

2

A generalization of Fermat’s little theorem

Let n, k ≥ 1 be positive integers, and put Sk (n) = 1k + 2k + · · · + nk . By writing the binomial theorem for the expressions (0 + 1)k = 1k 414

µ

¶ k (1 + 1) = 1 + · 1k−1 + . . . k−1 µ ¶ k k k (2 + 1) = 2 + · 2k−1 + . . . k−1 ... µ ¶ k k k (n + 1) = n + nk−1 + . . . , k−1 k

k

after addition, and reducing the common terms, we get the following well known formula: Lemma 1. One has the identity µ ¶ µ ¶ µ ¶ k k k k (n + 1) − (n + 1) = S1 (n) + S2 (n) + · · · + Sk−1 (n). 1 2 k−1 In what follows, we shall need also the following simple (but not so wellknown) lemma: µ ¶ µ ¶ m n−1 Lemma 2. If (m, k) = 1, then m| and k| . k k−1 Proof. The trivial identity m (m − 1) . . . (m − k + 1) m(m − 1) . . . (m − k + 1) = · 1 · 2...k k 1 · 2 . . . (k − 1) gives at once the combinatorial relation µ ¶ µ ¶ m m m−1 = . k k−1 k µ ¶ µ ¶ m m−1 This implies k =m . By assuming (m, k) = 1, since m dik k−1 vides the right µ hand ¶ side, by Euclid’s theorem (or Gauss’ µ lemma) ¶ it follows m m−1 that m divides on the left hand side. Similarly, k| . k k−1 µ ¶ p Corollary. If p is a prime number, then p| for all 1 ≤ k < p. k One has (p, k) = 1 and Lemma 2 applies. We now are able to prove the main result of this note: Theorem. Let (a, b) denote the g.c.d. of a and b. The following congruence is true: (n + 1)k − (n + 1) ≡ A(k, n) (mod k), 415

where A(k, n) =

X 1≤l≤k−1, (l,k)≥2

µ ¶ k Sl (n) l

and an empty sum is considered to µ be ¶ 0. k Proof. By Lemma 2 one has k| for all (k, l) = 1. By writing the sum l of Lemma 1 in two sums, namely the sum of terms with (k, l) = 1, and the otherone with (k, l) ≥ 2, we can write µ ¶ X k k (n + 1) − (n + 1) = M k + Sl (n) , M = integer, l 1≤l≤k−1, (l,k)≥2

and this implies the stated congruence. Remarks. 1) If k is prime, then by the Corollary one has A(k, n) = 0, so we can deduce the (”little”) Fermat theorem: (n + 1)k − (n + 1) ≡ 0

(mod k).

2) We have obtained also a new proof (without mathematical induction) of Fermat’s theorem. We note that this theorem has important applications e.g. in Cryotography (see e.g. [1]).

References [1] J. Sándor (in coop with B. Crstici), Handbook of number theory, II, Springer Verlag, 2005.

3

On an inequality of Klamkin

1. Introduction In 1974 M. S. Klamkin [3] proved the following result: Let x be a nonnegative real number, and m, n integers with m ≥ n ≥ 1. Then (m + n)(1 + xm ) ≥ 2n

1 − xm+n . 1 − xn

(1)

We note that for x = 1, the right side of (1) is understood as lim , when x→1

the inequality becomes an equality. Also, for x = 0 (1) becomes m + n ≥ 2n, which is true. For m = n there is equality in (1). In fact, it can be shown 416

that for all real numbers m > n > 0, and all x > 0, (1) holds true with strict inequality (see the solutions of (1) in [1]). Assume now that x = a ≥ 1, m = p, n = q, where p ≥ q ≥ 0 are real numbers. Then, since (1 + ap )(1 − aq ) = ap − aq + 1 − ap+q , after some transformations, (1) becomes equivalent to (p − q)(ap+q − 1) ≥ (p + q)(ap − aq ).

(2)

In the case of p − q ≤ 1, a weaker result than (2) appears in the famous monograph by D. S. Mitrinović [4] (3.6.26, page 276). For certain arithmetical applications of Klamkin’s inequality, see [5]. In what follows we will point out some surprising connections of inequality (2) (i.e., in fact (1)) with certain special means of two arguments. Also, a new application of (1) will be given.

2. Stolarsky means Let m, n > 0 and put p + q = m, p − q = n. Then p = and (2) gives

m+n m−n ,q= 2 2

am − 1 m > a(m−n)/2 . an − 1 n

By letting a =

(3)

x (x > y > 0), relation (3) may be written also as y µ

xm − y n n · xn − yn m

¶1/(m−n) >

√ xy.

(4)

If n = 1, the expression on the left side of (4) is called as the Stolarsky mean of x and y. Put µ S(m) = S(x, y, m) =

xm − y m 1 · x−y m

¶1/(m−1) .

It is not difficult to see that S can be defined also for all real numbers m 6∈ {0, 1}, while for m = 0, and m = 1, by the limits lim S(x, y, m) =

m→0

417

x−y ln x − ln y

and

1 lim S(x, y, m) = (y y /xx )1/(y−x) (y 6= x), m→1 e the definition of S can be extended to all real numbers m. Let x−y 1 L(x, y) = , I(x, y) = (y y /xx )1/(y−x) (x 6= y), ln x − ln y e L(x, x) = I(x, x) = x. These means are known as the logarithmic and identric means of x and y (see e.g. [8] for their properties). Stolarsky [10] has proved that S is a strictly increasing function of m. Therefore S(−1) < S(0) < S(1) < S(2), giving x+y √ . (5) xy < L(x, y) < I(x, y) < 2 Since S(−1) < S(0) < S(m) for m > 0, we get ¶ µ m x − y m 1/(m−1) √ xy < L(x, y) < , (6) m(x − y) which is an improvement of (4), when n = 1.

3. Main results We shall prove that the following refinement of (4) holds true: Theorem 1. µ m ¶ x − y m n 1/(m−n) √ m−n m−n 1/(m−n) xy < (L(x ,y )) < · (m > n). (7) xn − y n m ax − 1 f (p + q) (x > 0), where a > 1; and let ϕ(p) = x f (p − q) (p > q > 0), where q is fixed. We first show that ϕ is strictly increasing function. Since Proof. Put f (x) =

ϕ0 (p) =

f 0 (p + q)f (p − q) − f 0 (p − q)f (p + q) , f 2 (p − q)

f 0 (p + q) f 0 (p − q) > . Since p + q > p − q, f (p + q) f (p − q) this will follow, if f 0 /f = g is an increasing function. By

it will be sufficient to prove that

g 0 (t) = (f 0 (t)/f (t))0 =

f 00 (t)f (t) − (f 0 (t))2 , f 2 (t)

418

it will be sufficient to show that f is strictly log-convex (i.e. ln f is strictly convex). Lemma. The function f is strictly log-convex. Proof. After certain simple computations (which we omit here), it follows that tat ln t − (at − 1) , f 0 (t) = t2 f 00 (t) =

t2 at ln2 a − 2tat ln a + 2at − 2 , t3

and a2t − 2at − t2 at ln2 a + 1 t4 √ √ (at − 10 at ln at )(at − 1 + at ln at ) = . t4 √ √ h−1 √ Put at = h. Then h − 1 − h ln h > 0, since > h by L(h, 1) > h ln h (left side of (5)). This proves the log-convexity property of f for a > 1. Since ϕ is strictly increasing, one can write f 00 (t)f (t) − (f 0 (t))2 =

ϕ(p) >

lim

p→q,p>q

f (p + q)/f (p − q) =

a2q − 1 . 2q ln a

x , and the right side of (7) follows. y For the left side of (7) remark that again by the left side of (5) one has Write p + q = m, p − q = n, a =

L(xm−n , y m−n ) >

p xm−n y m−n = (xy)(m−n)/2 ,

which implies the desired inequality. Remark. ϕ being strictly increasing, it follows also that ap+q − 1 p − q · = a2q , p→∞ ap−q − 1 p + q

ϕ(p) < lim i.e.

(p − q)(ap+q − 1) ≤ (p + q)a2q (ap−q − 1), which is complementary to (2). 419

(8)

4. Arithmetical applications A divisor d of N is called unitary divisor of the positive integer N > 1, if (d, N/d) = 1. For k ≥ 0, let σk (N ) resp. σk∗ (N ) denote the sum of kth powers of divisors, resp. unitary divisors of N . Remark that σ0 (N ) = d(N ), σ0∗ (N ) = d∗ (N ) are the number of these divisors of N . It is well-known that (see e.g. [3], [9]) if the prime factorization of N is N=

r Y

pai i

i=1

(pi distinct primes, ai ≥ 1 integers), then σk (N ) =

r r Y Y k(a +1) (pi i − 1)/(pki − 1), d(N ) = (ai + 1), i=1

i=1

r Y ∗ i σk (N ) = (pka + 1), d∗ (N ) = 2r (= 2ω(N ) ), i

(9)

i=1

where ω(r) = r denotes the number of distinct prime divisors of N . Write now (1), and a reverse of it (see [1]) in the form 2n

xm+n − 1 xm+n − 1 m ≤ (m + n)(1 + x ) ≤ 2m , xn − 1 xn − 1

(10)

where x > 1, m ≥ n ≥ 1. Put n = k, m = kai , x = pi (i = 1, 2, . . . , r). Writing (10), after term-by-term multiplication, we get 2ω(N ) σk (N ) ≤ d(N )σk∗ (N ) ≤ 2ω(N ) β(N )σk (N ), where β(N ) =

r Y

(11)

ai (for this, and the other functions, too, see e.g. [6], [9]).

i=1

The left side of (11) appears also in [5]. Now, remarking that 2

ω(N )

r r Y Y β(N ) = (2ai ) ≤ 2ai = 2Ω(N ) , i=1

i=1

where Ω(N ) denotes the total number of prime factors of N (we have used the classical inequality 2a−1 ≥ a for all a ≥ 1), relation (11) implies also 2ω(N ) ≤

d(N )σk∗ (N ) ≤ 2Ω(N ) . σk (N ) 420

(12)

Theorem 2. The normal order of magnitude of log(d(N )σk∗ (N )/σk (N )) is (log 2)(log log N ). Proof. Let P be a property in the set of positive integers and set X ap (n) = 1 if n has the property P ; ap (n) = 0, otherwise. Let Ap (x) = ap (n). n≤x

If Ap (x) ∼ x (x → ∞) we say that the property P holds for almost all natural numbers. We say that the normal order of magnitude of the arithmetical function f (n) is the function g(n), if for each ε > 0, the inequality |f (n) − g(n)| < εg(n) holds true for almost all positive integers n. By a well-known result of Hardy and Ramanujan (see e.g. [2], [4], [6]), the normal order of magnitude of ω(N ) and Ω(N ) is log log N . By (12) we can write 1 (1 − ε)(log log N ) < ω(N ) ≤ log d(N )σk∗ (N )/σk (N ) log 2 ≤ Ω(N ) < (1 + ε) lg lg N for almost all N , so Theorem 2 follows. Acknowledgements. The author thanks Professor Klamkin for sending him a copy of [1] and for his interest in applications of his inequality.

References [1] J. M. Brown, M. S. Klamkin, B. Lepson, R. K. Meany, A. Stenger, P. Zwier, Solutions to Problem E2483, Amer. Math. Monthly, 82(1975), 758-760. [2] G. H. Hardy, E. M. Wright, An introduction to the theory of numbers, Oxford Univ. Press, 1960. [3] M. S. Klamkin, Problem E2483, Amer. Math. Monthly, 81(1974), 660. [4] E. Krätzel, Zahlentheorie, Berlin, 1981. [5] D. S. Mitrinović, Analytic inequalities, Springer Verlag, 1970. [6] D. S. Mitrinović, J. Sándor (in coop. with B. Crstici), Handbook of number theory, Kluwer Acad. Publ., 1995. [7] J. Sándor, On an inequality of Klamkin with arithmetical applications, Int. J. Math. E. Sci. Techn., 25(1994), 157-158. 421

[8] J. Sándor, On the identric and logarithmic means, Aequationes Math., 40(1990), 261-270. [9] J. Sándor (in coop. with B. Crstici), Handbook of number theory, II, Springer Verlag, 2005. [10] K. B. Stolarsky, The power and generalized logarithmic means, Amer. Math. Monthly, 87(1980), 545-548.

4

Euler-pretty numbers

Let ϕ be the Euler totient. For a fixed positive rational number k, we say that the pair (a, b) is k-Euler-pretty, if ϕ(a) =

b k

and ϕ(b) =

a . k

(1)

Our aim is to solve the system (1) for k = 2, i.e. to find all 2-Euler-pretty pairs: Theorem. All 2-Euler-pretty numbers are given by a = 2m , b = 2m , where m is an arbitrary positive integer. The proof of this result is based on the following auxiliary result. Lemma. For all positive integers u, v one has ϕ(uv) ≤ uϕ(v).

(2)

¶ Yµ ¶ Yµ ϕ(uv) 1 1 ϕ(v) Proof. = 1− ≤ 1− = , since for any prime uv p p v p|uv

p|v

1 divisor q - v one has 1 − ≤ 1. Thus (2) follows. There is equality only if each q prime divisor of u is a prime divisor of v, too. To prove the theorem, first remark that for k = 2, a and b must be even, so a = 2A, b = 2B. Then the system (1) becomes ϕ(2A) = B, ϕ(2B) = A.

(3)

Applying two times relation (2), one can write: B ≤ Aϕ(2) = A and A ≤ Bϕ(2) = B, since ϕ(2) = 1. Therefore B = A = x, and (3) becomes ϕ(2x) = x. 422

(4)

Apply again (2) for (4), one has: ϕ(2x) ≤ xϕ(2) = x, with equality only if each prime factor of x is also a prime factor of 2. In other words, x = 2s . Since ϕ(2) = 1, one may assume s ≥ 0. Thus a = 2s+1 = 2m , b = 2s+1 = 2m , where m ≥ 1. This finishes the proof of the theorem. Remark. By (3) we have solved in fact the following equation: ϕ(2ϕ(2A)) = A.

(5)

The more general system (1) can be reduced (for positive integers k), to a similar equation, too, namely ϕ(kϕ(kA)) = A.

(6)

What are the most general solutions of this equation? An equation of type (4) has been studied, by other arguments, in [1].

References [1] J. Sándor, On the Open Problem PD.90 (Romanian), Gamma, 11(1988), no.1-2, 26-27.

5

Abundant numbers involving the smallest and largest prime factors

Problem 11 of [1] asks for abundant numbers n such that p(n) + P (n) is also abundant. Here p(n), resp. P (n) denote the smallest, resp. largest prime factors of n. Recall that n is called abundant, if σ(n) > 2n, where σ(n) denotes the sum of divisors of n. Our aim is to prove the following results: Theorem 1. Let n = 51·5b ·7c , where b ≥ 2 and c ≥ 1 are positive integers. Then p(n) + P (n) and n are abundant at the same time. Proof. Since 51 = 3 · 17, clearly p(n) + P (n) = 3 + 17 = 20, an abundant number: σ(20) = σ(4) · σ(5) = 42 > 40. On the other hand, by the multiplicatively property of σ one can write σ(n) = 4 · 18

5b+1 − 1 7c+1 − 1 · > 2 · 51 · 5b · 7c 4 6

iff (5b+1 − 1)(7c+1 − 1) > 34 · 5b · 7c , or 5b · 7c − 5b+1 − 7c+1 + 1 > 0. Equivalently, (5b − 7)(7c − 5) > 34 423

(1)

Now, in (1) 7c − 5 ≥ 2 for any c ≥ 1 and 5b − 7 ≥ 18 for any b ≥ 2. Since 2 · 18 = 36 > 34, relation (1) is true. Remark 1. Therefore for b = 2, 3, . . . , c = 1, 2, . . . we get such numbers as n = 8925, 44625, 62475, 223125, . . . Problem 12 of [1] asks for abundant numbers n such that the sum of all composite numbers strictly between p(n) and P (n) is also abundant. The following is true: Theorem 2. For any a ≥ 1, b ≥ 1, the number n = 19 · 2a · 3b satisfies the required property. Proof. Since p(n) = 2, P (n) = 19, the sum of the composite numbers between 3 and 18 is 4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 112. Since σ(112) = 248 > 224, this is abundant number. Similarly, µ a+1

σ(n) = (2

− 1)

3b+1 − 1 2

¶ · 20 > 2 · 2a · 3b · 19

iff 5(2a+1 ·3b+1 −2a+1 −3b+1 +1) > 19·2a ·3b , or 11·2a ·3b −10·2a −15·3b +5 > 0. This can be written also as 2a ·(11·3b −10) > 15·3b −5. Now 2a ·(11·3b −10) ≥ 2 · (11 · 3b − 10) = 22 · 3b − 20 > 15 · 3b − 5 by 7 · 3b > 15, which is true for any b ≥ 1, since 7 · 3b ≥ 7 · 3 = 21. Remark 2. For a = 1, 2, . . . , b = 1, 2, . . . we get various numbers n = 114, 228, 342, 456, 684, . . .


6

An inequality with sh x and ch x We will determine all α, β, γ > 0 such that µ

sh x x

¶β

µ +

Here sh x =

sh x x

¶γ

≤ (ch x + 1)α for all x ∈ R.

ex − e−x x x3 x5 = + + + ... 2 1! 3! 5! 424

(1)

and

ex + e−x x2 x4 =1+ + + ..., 2 2! 4! where we have used the classical formula ch s =

et = 1 +

t t2 + + ... 1! 2!

Now (1) for β = γ = α = 1 can be written as sh x x2 x4 ch x + 1 x2 x4 =1+ + + ··· ≤ =1+ + + ... x 3! 5! 2 2 · 2! 2 · 4!

(2)

sh 0 = 1 since lim shxx = 1). Now (2) trivially holds true for x = 0, while for x→0 0 x 6= 0 remark that 1 1 1 1 < , < ,... 3! 2 · 2! 5! 2 · 4!

(

sh x ≥ 1, for all x ∈ R, b = ch x + 1 ≥ 2. For x = 0 relation x α (2) gives 2 ≤ 2 , so α ≥ 1. By (2) one has b ≥ 2a. Clearly bα ≥ (2a)α so if β ≤ α, γ ≤ α, then aβ−α + aγ−α ≤ 2 ≤ 2α by α ≥ 1. Therefore, inequality (1) holds true for all α ≥ 1 and all 0 < β ≤ α, 0 < γ ≤ α. If, for example, β > α, then one can obtain a > 1, b > 2a so that aβ + aγ > bα (indeed, for ln b , one has β ln a > α ln b, so for β > α(1 + ε), ε > 0 the inequality β >α ln a is valid). In the same manner, γ > α implies the reverse inequality for some sh t , t > 0 and g(t) = ch t + 1, t > 0, f : (0, ∞) → (1, ∞), a, b. Since f (t) = t g : (0, ∞) → (2, ∞) are surjective functions, all solutions of (1) are Now, let a =

α ≥ 1,

7

0 < β ≤ α,

0 < γ ≤ α.

(3)

On geometric numbers

1. Introduction

³ n´ Let d be a positive divisor of the integer n > 1. If d, = 1, then d is d called a unitary divisor of n. If the greatest common unitary divisor of d and n/d is 1, then d is called a bi-unitary divisor of n. If n = pa11 . . . par r > 1 is the prime factorization of n, a divisor d of n is called an exponential divisor (or e-divisor, for short), if d = pb11 . . . pbrr , with bi |ai (i = 1, r). For the history 425

of these notions, as well as the connected arithmetical functions, see e.g. [2], [7]. Let T (n), T ∗ (n), T ∗∗ , resp. Te (n) denote the product of divisors, unitary divisors, bi-unitary divisors, resp. e-divisors of n. It½ is easily seen ¾ that, if n n ,..., , implying d1 , . . . , ds are all divisors of n, then {d1 , . . . , ds } = d1 ds n n that d1 . . . ds = . . . , giving d1 ds T (n) = nd(n)/2 ,

(1)

where s = d(n) is the number of divisors of n. If {d∗1 , . .½. , d∗m } are all¾unitary n n , . . . , ∗ , so, in divisors of n, then one has similarly, {d∗1 , . . . , d∗m } = ∗ d1 dm analogy with (1), one has ∗ T ∗ (n) = nd (n)/2 , (2) where m = d∗ (n) denotes the number of unitary divisors of n. In a completely similar manner, one can write ∗∗ (n)/2

T ∗∗ (n) = nd

,

(3)

where d∗∗ (n) denotes the number of bi-unitary divisors of n. It is well known, that if p = paa1 . . . par r > 1 is the prime representation of n, then d(n) = (a1 + 1) . . . (ar + 1), d∗ (n) = 2r , d∗∗ (n) =

Y ai even

ai

Y

(aj + 1). (4)

aj odd

For Te (n), the things are slightly more complicated, for the following formula holds true (see [6]): σ(a1 )d(a2 )...d(ar )

Te (n) = p1 where σ(k) =

X

r )d(a1 )...d(ar−1 ) . . . pσ(a , r

(5)

d is the sum of (ordinary) divisors of k.

d|k

Let σ ∗ (k), resp. σ ∗∗ (k) denote the sum of unitary, resp. bi-unitary divisors of k. In 1948 O. Ore [3] has studied the classical means related to the divisors of an integer. Let A(n), G(n), H(n) denote the arithmetic, geometric, and harmonic means of all ordinary divisors of n, i.e. A(n) =

d1 + · · · + ds , s 426

G(n) = (d1 . . . ds )1/s ,

µ H(n) = s/

¶

1 1 + ··· + d1 ds

.

It is immediate from the above notations, that A(n) =

σ(n) , d(n)

G(n) = (T (n))1/d(n) ,

where the last formula is a consequence of

H(n) =

X1 d|n

=

d

nd(n) , σ(n)

(6)

1X d. n d|n

Ore called a number n arithmetic, geometric, resp. harmonic, if A(n), G(n), resp. H(n) is integer. Similarly, other authors have studied the unitary analogues A∗ (n) =

σ ∗ (n) , d∗ (n)

∗ (n)

G∗ (n) = (T ∗ (n))1/d

,

H ∗ (n) =

nd∗ (n) . σ ∗ (n)

(7)

Recently, the present author [5] has studied the bi-unitary harmonic numbers n, i.e. with H ∗∗ (n) an integer; where H ∗∗ (n) is the third of the following means: A∗∗ (n) =

σ ∗∗ (n) , d∗∗ (n)

∗∗ (n)

G∗∗ (n) = (T ∗∗ (n))1/d

,

H ∗∗ (n) =

nd∗∗ (n) . σ ∗∗ (n)

(8)

He also introduced some variants of these bi-unitary harmonic numbers, e.g. H1 (n) =

nd(n) , σ ∗ (n)

H2 (n) =

nd∗ (n) , etc. σ(n)

(four other related fractions, involving also d∗∗ (n), σ ∗∗ (n)). It is immediate that, by (1)-(3), one has G(n) = G∗ (n) = G∗∗ (n) =

√ n,

(9)

so a geometric number, etc., is in fact a perfect square. We note that actually the relationships of arithmetic, geometric and harmonic numbers is not perfectly known, for example, it is conjectured (but not proved up to now, see e.g. [1]) that no harmonic number is geometric (i.e., a perfect square). Many harmonic numbers seem to be arithmetic, too; but their proportion is not clarified (see e.g. [2], [7]). Our aim in what follows is to introduce and study certain notions of geometric numbers, which are not so obvious than the ones given by (9). 427

2. Main notions and results The geometric mean of e-divisors is clearly given by Ge (n) = (Te (n))1/de (n) , and since it is known that de (n) = d(a1 ) . . . d(ar ) (see e.g. [7]), then by (5) one can write: σ(a )/d(a1 ) r )/d(ar ) Ge (n) = p1 1 . . . pσ(a . (10) r Let us now introduce the following expressions: ∗ (n)

G1 (n) = (T ∗ (n))1/d(n) ,

G2 (n) = (T (n))1/d

G3 (n) = (T ∗∗ (n))1/d(n) ,

G4 (n) = (T (n))1/d

∗∗ (n)

G5 (n) = (T ∗ (n))1/d

,

,

∗∗ (n)

,

∗ (n)

G6 (n) = (T ∗∗ (n))1/d

.

(11)

We note that, we could introduce also G1e (n) = (T ∗ (n))1/de (n) , etc., but these expressions will not be considered here. In what follows ω(n) = r will denote the number of distinct prime factors of n. We say that n is G1 -geometric (or G1 -number), if G1 (n) is an integer. The G2 -, etc. numbers are defined in a similar manner. We say that n is e-geometric, if Ge (n) is an integer. Theorem 1. Let n = pa11 . . . par r > 1 be the canonical factorization of n. Then n is e-geometric number if and only if all of a1 , . . . , ar are (ordinary) arithmetic numbers. Proof. The proof is a consequence of (10); the definition of arithmetic numbers, and the following auxiliary result: Lemma 1. Let b = pα1 1 . . . pαr r > 1, where pi are distinct primes, and αi are positive rational numbers. Then n is an integer if and only if all of α1 , . . . , αr are integers. b1 br Proof of the Lemma. Let us write α1 = , . . . , αr = , with common k k nominator k, and b1 , . . . , br positive integers. Then nk = pb11 . . . pbrr , so n cannot have other prime divisors, then p1 , . . . , pr . Indeed, if p|nk , (i.e. p|n), then p mr be 1 divides one of pb11 , . . . , pbrr , so p is one of p1 , . . . , pr . Let n = pm 1 . . . pr the prime factorization of n. Then by the unique factorization theorem one b1 br must have km1 = b1 , . . . , kmr = br , so = α1 , . . . , = αr are the integers k k m1 , . . . , mr . Lemma 1 clearly implies Theorem 1. 428

Remark. Two notions of e-harmonic numbers have been recently introduced by the present author in [5]. Lemma 2. One has, for all n ≥ 1: d∗ (n)|d∗∗ (n)

and

d∗∗ (n) ≤ d(n).

(12)

Proof. Let us suppose that there are t even numbers, and q odd num∗∗ bers Y between Y a1 , . . . , ar (t, q ≥ 0, t + q = r). Then, clearly d (n) = t q t+q r ∗ ai (aj +1) is divisible by 2 2 = 2 = 2 = d (n). This proved the ai even

aj odd

divisibility relation of (12). The second relation is trivial, by remaking that for ai even one has ai < ai + 1. It is immediate that in the first relation one has equality only if n has the form n = pε11 . . . pεrr , where εi ∈ {1, 2} for all i = 1, r. There is equality in the second relation of (12) only if all of ai are odd (in which case, one has also σ ∗∗ (n) = σ(n), so A∗∗ (n) = A(n), H ∗∗ (n) = H(n), too). Corollary. d∗ (n) ≤ d∗∗ (n) ≤ d(n) for all n ≥ 1. Theorem 2. For n > 1, there are no G1 -numbers n. The general form of G5 -numbers n is n = p21 . . . p2r , where pi , i = 1, r, are distinct primes. ∗ α d∗ (n)/2d(n) α d∗ (n)/2d(n) Proof. Since G1 (n) = nd (n)/2d(n) = p1 1 . . . pr r , by Lemma 1, one must have 2d(n)|α1 d∗ (n), etc., i.e. 2(α1 + 1) . . . (αr + 1)|α1 · 2r . But (α2 + 1) . . . (αr + 1) ≥ 2r−1 , and α1 + 1 > α1 , thus 2(α1 + 1) . . . (αr + 1) > α1 · 2r , in contradiction with the above divisibility relation. This proves the first part of Theorem 2. For the second part, note that, since G5 (n) = nd

∗ (n)/2d∗∗ (n)

α d∗ (n)/2d∗∗ (n)

= p1 1

. . . pαr r d

∗ (n)/2d∗∗ (n)

,

we must have Y Y Y Y 2 αi (αj + 1)|α1 · 2r , . . . , 2 αi (αj + 1)|αr · 2r , where αi are the even, while αj the odd exponents. Now, remark that if α1 is odd, then the left side is divisible by 2r+1 , but the right side only by 2r , a contradiction. So α1 , . . . , αr are all even. Since 2α1 . . . αr |α1 2r implies α2 . . . αr |2r−1 , and since 2r−1 |α2 . . . αr , we must have α2 = · · · = αr = 1. Similarly, α1 = 1, and this finishes the proof of the second part of Theorem 2. Theorem 3. n > 1 is a G6 -number if and only if a prime power pα kn satisfies one of the following conditions: a) α is even; 429

b) α ≡ −1 (mod 4); c) α is odd, α 6≡ −1 (mod 4), d∗∗ (n/(α + 1)) ≡ 0 (mod 2ω(n) ). The number n > 1 is a G2 -number, if and only if αd(n) ≡ 0

(mod 2ω(n)+1 ).

∗∗ (n)/2d∗ (n)

Proof. G6 (n) = nd

, so by Lemma 1 we must have Y Y 2r+1 |α αi (αj + 1),

where αi are the even, and αj are the odd exponents. In the cases a) and b) these are always true, by Lemma 2. Now, case c) follows by a direct verification. ∗ Since G2 (n) = nd(n)/2d (n) , by Lemma 1 we must have 2r+1 |αd(n), where r = ω(n). Corollary. Any ordinary geometric number if a G6 number. Any number n = pα1 1 . . . pαr r with αi ≡ −1 (mod 4), i = 1, r, is a G2 -number. Indeed, if n is perfect square, then all α are even, and case a) of Theorem 3 applies. If 4|(α + 1), then clearly 4r |d(n), and 2r+1 |4r for any r ≥ 1 (r = ω(n)) implies the result. Theorem 4. There are no G4 -numbers n > 1. ∗∗ Proof. Let n = pα1 1 . . . pαr r > 1. Then G4 (n) = nd(n)/2d (n) , so by Lemma 1, n is a G4 -number iff 2d∗∗ (n)|αd(n) for all α. Remark that if all α are odd, then d∗∗ (n) = d(n), so 2|α, and this is impossible. Therefore, there exists αi = even. By relation (4), one obtains Y Y 2 αi |α (αi + 1) for all α. Particularly, when Y α is one Yof αi , this is impossible. Indeed, by reducing with α, one obtains 2 αi | (αi + 1), where the left side is even, while the αi 6=α

right side odd. This gives a desired contradiction. Theorem 5. Let n = pα1 1 . . . pαr r > 1. Then n is a G3 -number iff for any prime power pα kn one has Y Y 2 (αi + 1)|α αi , (∗) where αi denote the even exponents (here an empty product is taken to be 1). ∗∗ Proof. G3 (n) = nd (n)/2d(n) , so 2d(n)|αd∗∗ (n). By using relation (4), this reduces to (∗). 430

Remarks. If all α are odd, then (∗) reduces to 2|α, which is impossible. If r = 1, then n = pα with α even, so 2(α+1)|α2 , which cannot be true for any α, 2b+1 2b since (α+1, α2 ) = 1. If r = 2, then n = p2a or n = p2a 1 p2 1 p2 . In the first case (∗) becomes 2(2a + 1)|α(2a) for α = 2a and α = 2b + 1. But 2(2a + 1)|(2a)2 is not true by (2a + 1, (2a)2 ) = 1. In the second case, 2(2a + 1)(2b + 1)|α(2a)(2b) for α = 2a and α = 2b. Here 2(2a + 1)(2b + 1)|2a(2a)(2b) by (2a + 1, 4a2 ) = 1 implies (2a+1)|b; and similarly (2b+1, 4b2 ) = 1 gives (2b+1)|a. Since b ≥ 2a+1 and a ≥ 2b + 1 ≥ 4a + 3, this is an obvious contradiction. Therefore, for all G3 -numbers n one has ω(n) ≥ 3.

References [1] G.L. Cohen, R.M. Sordy, Harmonic seeds, Fib. Quart. 36(1998), 386390. [2] R.K. Guy, Unsolved problems in number theory, Third ed., Springer Verlag, 2004. [3] O. Ore, On the averages of the divisors of a number, Amer. Math. Monthly, 55(1948), 615-619. [4] J. Sándor, On bi-unitary harmonic numbers, submitted. [5] J. Sándor, On exponentially harmonic numbers, submitted. [6] J. Sándor, On multiplicatively e-perfect numbers. J. Ineq. Pure Appl. Math., 5(2004), no.4, article 114 (electronic). [7] J. Sándor, Handbook of number theory, II, Springer Verlag, 2004.

8

The sum-of-divisors minimum and maximum functions

1. Let f : N∗ → N be a given arithmetic function, and A ⊂ N∗ a given set. The arithmetic function FfA (n) = min{k ∈ A : n|f (k)}

(1)

has been introduced in [7] and [6]. For A = N∗ , f (k) = k! one obtains the Smarandache function; for A = N∗ , A = P = {2, 3, 5, . . . } = set of all primes, one obtains a function P (n) = min{k ∈ P : n|k!} 431

(2)

For properties of this function, see [7], [6]. For A = {k 2 : k ∈ N∗ } = set of perfect squares, and f (k) = k! one obtains the function Q(n) = min{m2 ≥ 1 : n|(m2 )!}, (3) while for A = set of squarefree numbers ≥ 1, f (k) = k! we get Q1 (n) = min{m ≥ 1 squarefree: n|m!}

(4)

For properties of Q(n) and Q1 (n), see [11]. The ”dual” function of (1) has been defined by GA g (n) = max{k ∈ A : g(k)|n}

(5)

where g : N∗ → N is a given function. Particularly for A = N∗ , g(k) = k! one obtains the dual of the Smarandache function S∗ (n) = max{k ≥ 1 : k!|n}

(6)

For properties of this function, see [7], [6]. F. Luca [4], K. Atanassov [1] and M. Le [2] have proved in the affirmative a conjecture of the author stated in [7] and [6]. For A = N∗ , f (k) = g(k) = ϕ(k) (where ϕ is Euler’s totient) in (1), resp. (5) one obtains the Euler minimum, resp. maximum-functions, defined by E(n) = min{k ≥ 1 : n|ϕ(k)},

(7)

E∗ (n) = max{k ≥ 1 : ϕ(k)|n}

(8)

For properties of these functions, see [5], [8]. When A = N∗ , f (k) = d(k) = number of divisors of k, one has the divisor minimum function (see [7], [6], [9]): D(n) = min{k ≥ 1 : n|d(k)}

(9)

It is interesting to note that the divisor maximum function (i.e. the ”dual” of D(n)) given by D∗ (n) = max{k ≥ 1 : d(k)|n} (10) is not well-defined! Indeed, for any prime p we have d(pn−1 ) = n, and pn−1 is unbounded as p → ∞. When A is a finite set, however, D∗A (n) = max{k ∈ A : d(k)|n} 432

(11)

does exist. When A = N∗ , f (k) = g(k) = S(k) = min{m ≥ 1 : k|m!} (Smarandache function) one obtains the Smarandache minimum and maximum functions, given by Smin (n) = min{k ≥ 1 : n|S(k)}, (12) Smax (n) = max{k ≥ 1 : S(k)|n}.

(13)

These functions have been introduced and studied recently in [10]. 2. Let σ(n) be the sum of divisors of n. The function Σ(n) = min{k ≥ 1 : n|σ(k)}

(14)

has been introduced in [7], [6] (denoted there by Fσ ). Let k be a prime of the form k = an − 1, where n ≥ 1 is given. By Dirichlet’s theorem on arithmetical progressions, such a prime does exist. Then clearly σ(k) = an, so n|σ(k), and Σ(n) is well defined. The dual of Σ(n) is Σ∗ (n) = max{k ≥ 1 : σ(k)|n}

(15)

Since σ(1) = 1|n and σ(k) ≥ k, clearly Σ∗ (n) ≤ n, so this function is correctly defined. The aim of this note is the initial study of these functions Σ(n) and Σ∗ (n). Some values of Σ(n) are: Σ(1) = 1, Σ(2) = 3, Σ(3) = 2, Σ(4) = 3, Σ(5) = 8, Σ(6) = 5, Σ(7) = 4, Σ(8) = 7, Σ(9) = 10, Σ(11) = 43, Σ(12) = 6, Σ(13) = 9, Σ(14) = 12, Σ(15) = 8, Σ(16) = 21, Σ(17) = 67, Σ(18) = 10, Σ(19) = 37, Σ(20) = 19, Σ(21) = 20, Σ(22) = 43, Σ(23) = 137, Σ(24) = 14, Σ(25) = 149, Σ(26) = 45, Σ(27) = 34, Σ(28) = 12, Σ∗ (1) = 1, Σ∗ (2) = 1, Σ∗ (3) = 2, Σ∗ (4) = 3, Σ∗ (5) = 1, Σ∗ (6) = 5, Σ∗ (7) = 4, Σ∗ (8) = 7, Σ∗ (9) = 2, Σ∗ (10) = 1, Σ∗ (11) = 1, Σ∗ (12) = 11, Σ∗ (13) = 9, Σ∗ (14) = 13, Σ∗ (15) = 8, Σ∗ (16) = 7, Σ∗ (17) = 1, Σ∗ (18) = 17, Σ∗ (19) = 1, Σ∗ (20) = 19, Σ∗ (21) = 4, Σ∗ (22) = 1, Σ∗ (23) = 1, Σ∗ (24) = 23, Σ∗ (25) = 1, Σ∗ (26) = 9, Σ∗ (27) = 2, Σ∗ (28) = 12. 3. The first theoretical result gives informations on values of these functions at n = p + 1, where p is a prime: Theorem 1. If p is a prime, then Σ(p + 1) ≤ p ≤ Σ∗ (p + 1)

(16)

Proof. Since (p+1)|σ(p) = p+1, by definition (14) one can write Σ(p+1) ≤ p. Similarly, definition (15) gives (by σ(p) = (p + 1)|(p + 1)) Σ∗ (p + 1) ≥ p. 433

Remark. On the left side of (16) one can have equality, e.g. Σ(3) = 2, Σ(6) = 5, Σ(8) = 7. But the inequality can be strict, as Σ(12) = 6 < 11, Σ(18) = 10 < 17. For the right side of (16) however, one can prove the more precise result: Theorem 2. For all primes p, one has Σ∗ (p + 1) = p

(17)

Proof. First we prove that for all n ≥ 2 we have Σ∗ (n) ≤ n − 1

(18)

Indeed, since σ(k)|n, clearly we must have σ(k) ≤ n. On the other hand, for all k ≥ 2 we have σ(k) ≥ k + 1 (with equality only for k = prime), so k ≤ n − 1, and this is true for all k, so (18) follows. Let now n = p + 1 ≥ 3 in (18). Then Σ∗ (p + 1) ≤ p, which combined with (16) implies relation (17). Theorem 3. Let p be a prime and suppose that (p + 1)|n

(19)

Σ∗ (n) ≥ p

(20)

Then

Proof. Indeed, by σ(p) = (p + 1)|n, and definition (15), relation (20) follows. By letting p = 2, 3, 5, 7, 11 one gets: Corollary. If 3|n, then Σ∗ (n) ≥ 2. (21) If 4|n, then Σ∗ (n) ≥ 3. (22) If 6|n, then Σ∗ (n) ≥ 5. (23) If 8|n, then Σ∗ (n) ≥ 7. (24) If 12|n, then Σ∗ (n) ≥ 11. (25) Remark. If 7|n, then Σ∗ (n) ≥ 4. (26) Indeed, σ(4) = 7|n. If 15|n, then Σ∗ (n) ≥ 8. (27) Indeed, σ(8) = 15|n. It is immediate that Σ(n) = 1 only for n = 1. On the other hand, there exist many integers m with Σ∗ (m) = 1. Theorem 4. Let p be a prime such that p 6∈ σ(N∗ ) 434

(28)

Then Σ∗ (p) = 1

(29)

Proof. Remark that σ(k)|p ⇔ σ(k) = 1 or σ(k) = p. Now, if (28) is true, then the equation σ(k) = p is impossible for all k ≥ 1, so σ(k) = 1, i.e. k = 1, giving relation (29). For example, p = 17, 19, 23 satisfy relation (28). Theorem 5. If for all d > 1, d|n one has d 6∈ σ(N∗ ),

(30)

Σ∗ (n) = 1

(31)

then

Proof. Let d > 1, d|n. If d 6∈ σ(N∗ ), then the equation σ(k) = d is impossible. But then σ(k)|n is also impossible for σ(k) > 1, yielding (31). For example, n = 10, 22, 25 satisfy relation (30). Theorem 6. Let n be odd and suppose that Σ∗ (n) 6= 1, 2. Then µ Σ∗ (n) ≤

−1 +

√ ¶2 −3 + 4n 2

(32)

Proof. We use the following well-known results: Lemma 1. σ(k) is odd iff k = m2 or k = 2α m2 , where α ≥ 1 and m is an odd integer. (33) α1 α r Proof. Let k = p1 . . . pr . Then σ(k) = (1 + p1 + · · · + pα1 1 ) . . . (1 + pr + · · · + pαr r ). If k is odd, the σ(k) is odd if each term 1 + p1 + · · · + pα1 1 , . . . , 1 + pr + · · · + pαr r is odd, and since pi (1 = 1, r) are all odd numbers, we must have α1 = even, . . . , αr = even. This gives k = m2 , with m = odd. When k is even, then k = 2α pα1 1 . . . pαr r , and since σ(2α ) = 2α + 1 = odd, by the same argument as above, k = 2α m2 , with m = odd. Lemma 2. If k is composite, then √ σ(k) ≥ k + k + 1 (34)

435

√ Proof. Write k = ab, where√1 < a ≤ b < k. Then k ≤ b2 , so b ≥ k, implying σ(k) ≥ 1 + b + k ≥ 1 + k + k, i.e. relation (34). When k = p2 , with p an odd prime, one has equality since σ(p2 ) = p2 + p + 1. Now, if σ(k)|n and n is odd, then clearly σ(k) must be odd, too. Now, by (33) this is possible only when k = m2 or k = 2α m2 , with m ≥ 1 odd. If m > 1, then k = m2 is composite, while if m = 1 in k = 2α m2 , then k = 2α is prime only if α = 1, i.e. √ if k = 2. Supposing k 6= 1, 2 then √ k is always composite, so σ(k) √ ≥ k + k + 1. Since σ(1) ≤ n, we get k + k + 1 − n ≤ 0 √ −1 + −3 + 4n so k ≤ , and this gives (32). 2 Remark. For example, by (26), for 7|n, n odd, (32) is true. Theorem 7. If n ≥ 4, then Σ(n) ≥ 3. For all n ≥ 4, Σ(n) > n2/3

(35)

Proof. Σ(n) = 1 iff n|1, when n = 1. For Σ(n) = 2 we have σ(2) = 3 so n|3 ⇔ n = 1, 3. Thus for n ≥ 4, we have k = Σ(n) ≥ 3. Now, if n|σ(k), then clearly n ≤ σ(k). Let k ≥ 3. Then, it is known (see [3]) that √ (36) σ(k) < k k √ By n < k k = k 3/2 , inequality (35) follows. Corollary. For all m ≥ 2 (left side), and m ≥ 1 (right side): (2m+1 − 1)2/3 < Σ(2m+1 − 1) ≤ 2m

(37)

Proof. 2m+1 − 1 > 4 for m ≥ 2, and the left side is a consequence of (35). Now, the right side follows by (2m+1 − 1)|σ(2m ), since σ(2m ) = 2m+1 − 1, and apply definition (14). Theorem 8. Let f : [1, ∞) → [1, ∞) be given by f (x) = x + x log x. Then for all n ≥ 1, Σ(n) ≥ f −1 (n), (38) where f −1 is the inverse function of f . X Xn X1 X 1 Proof. σ(n) = d= =n ≤n ≤ n(1 + log n) as it is d d d d|n

d|n

d|n

1≤d≤n

1 1 well known that 1 + + · · · + ≤ 1 + log n for all n ≥ 1. Thus if n|σ(k), then 2 n n ≤ σ(k) ≤ f (k), so (38) follows. The function f is strictly increasing and continuous, so it is bijective, having an inverse function f −1 : [1, ∞) → [1, ∞). 436

√ √ Remark. The inequality f (x) < x x, i.e. log x < x − 1 is true for √ x sufficiently large (e.g. x ≥ e3 ). Indeed, let g(x) = x −√lg x − 1, when x−2 g(e3 ) = e3/2 − 4 > 0 by e3 ≈ 19.6 > 42 = 16, and g 0 (x) = > 0 for 2x √ x > 4. So g(x) ≥ g(e3 ) > 0 for x ≥ e3 . Thus x + x lg x < x x. By putting x = n2/3 we get f (n2/3 ) < n, i.e. for n2/3 ≥ e3 (m ≥ e9/2 ) we get: f −1 (n) > n2/3 for n ≥ e9/2

(39)

which improves, by (38), inequality (35). For values of Σ(n) and Σ∗ (n) at primes n = p the following is true: Theorem 9. For all primes p ≥ 5, 1 ≤ Σ∗ (p) ≤ p − 2 and

µ Σ∗ (p) ≤

−1 +

¶2 √ −3 + 4p 2

(40)

(41)

Proof. The inequality Σ∗ (n) ≥ 1 is true for all n (but remains an Open Problem the determination of all n with equality). Now, remark that σ(k)|p iff σ(k) = 1 or σ(k) = p. If σ(k) > 1, then by σ(k) ≥ k+1 we get k ≤ p−1. But we cannot have equality, since then k = q = prime, when σ(q) = q+1 = p ≥ 5 and this is impossible, since q + 1 is even for q ≥ 3, while for q = 2, q + 1 =√3 < 5. Thus k ≤ p − 2, so (40) follows. By applying the inequality σ(k) ≥ k + k + 1 (see (34)) then one arrives at (41), which is sharp, since e.g. Σ∗ (7) = 4 ≤ 4. Theorem 10. For all Mersenne primes p one has Σ(p) ≤

p+1 2

(42)

Proof. This follows from the right side of (37), by remarking that when p+1 p = 2m+1 − 1 is a prime, by Σ(2m+1 − 1) ≤ 2m = we get (42). 2

References [1] K. T. Atanassov, Remark on J´ ozsef S´ andor and Florian Luca’s theorem, C. R. Acad. Bulg. Sci. 55(2002), no. 10, 9-14. [2] M. Le, A conjecture concerning the Smarandache dual function, Smarandache Notion J. 14(2004), 153-155. 437

[3] C. C. Lindner, Problem E1888, Amer. Math. Monthly 73(1966), Solution by A. Bager and S. Russ, same journal 74(1967), 1143. [4] F. Luca, On a divisibility property involving factorials, C. R. Acad. Bulg. Sci. 53(2000), no. 6, 35-38. [5] P. Moree, H. Roskam, On an arithmetical function related to Euler’s totient and the discriminator, Fib. Quart. 33(1995), 332-340. [6] J. Sándor, On certain generalizations of the Smarandache function, Smarandache Notions J. 11(2000), no. 1-3, 202-212. [7] J. Sándor, On certain generalizations of the Smarandache function, Notes Number Theory Discr. Math. 5(1999), no. 2, 41-51. [8] J. Sándor, On the Euler minimum and maximum functions, (to appear). [9] J. Sándor, A note on the divisor minimum function, (to appear). [10] J. Sándor, The Smarandache minimum and maximum functions, (to appear). [11] J. Sándor, The Smarandache function of a set, Octogon Math. Mag. 9(2001), No. 1B, 369-371.

9

On certain new means and their Ky Fan type inequalities

1. Introduction Let x = (x1 , . . . , xn ) be an n-tuple of positive numbers. The unweighted arithmetic, geometric and harmonic means of x, denoted by A = An , G = Gn , H = Hn , respectively, are defined as follows n

1X xi , A= n i=1

G=

Ã n Y

!1/n xi

i=1

Ã

,

n X 1 H = n/ xi

! .

i=1

Assume 0 < xi < 1, 1 ≤ i ≤ n and define x0 := 1 − x = (1 − x1 , . . . , 1 − xn ). Throughout the sequel the symbols A0 = A0n , G0 = G0n and H 0 = Hn0 will stand for the unweighted arithmetic, geometric and harmonic means of x0 . The arithmetic-geometric mean inequality Gn ≤ An (and its weighted variant) played an important role in the development of the theory of inequalities. 438

Because of its importance, many proofs and refinements have been published. The following inequality is due to Ky Fan: ¸ µ remarkable 1 If xi ∈ 0, (1 ≤ i ≤ n), then 2 G A ≤ 0 0 G A

(1)

with equality only if x1 = · · · = xn . The paper by H. Alzer [1] (who obtained many results related to (1)) contains a very good account up to 1995 of the Ky Fan type results (1). For example, in 1984 Wang and Wang [11] established the following counterpart of (1): H G ≤ 0 H0 G

(2)

1 x2 x1 1/(x2 −x1 ) (x /x1 ) (x1 6= x2 ), I(x, x) = x denote e 2 the so-called identric mean of x1 , x2 > 0. In 1990 J. Sándor [8] proved the following refinement of (1) in the case of two arguments (i.e. n = 2): Let I = I(x1 , x2 ) =

G I A ≤ 0 ≤ 0, 0 G I A

(3)

where I 0 = I 0 (x1 , x2 ) = I(1 − x1 , 1 − x2 ). We note that, inequality (14) in Rooin’s paper [6] is exactly (3). In 1999 Sándor and Trif [10] have introduced an extension of the identric mean to n arguments, as follows. For n ≥ 2, let En−1 = {(λ1 , . . . , λn−1 ) : λi ≥ 0, 1 ≤ i ≤ n − 1, λ1 + · · · + λn−1 ≤ 1} be the Euclidean simplex. Given any probability measure µ on En−1 , for a continuous strictly monotone function f : (0, ∞) → R, the following functional means of n arguments can be introduced: ÃZ Mf (x; µ) = f −1

where xλ =

n X

! f (xλ)dµ(λ) ,

(4)

En−1

xi λi denotes the scalar product, λ = (λ1 , . . . , λn−1 ) ∈ En−1 ,

i=1

and λn = 1 − λ1 − · · · − λn−1 . 439

For µ = (n − 1)! and f (t) = 1/t, one obtains the unweighted logarithmic mean, studied by A.O. Pittenger [5]. For f (t) = ln t, however we obtain a mean ÃZ ! I = I(x) = exp

ln(xλ)dµ(λ)

(5)

En−1

which may be considered as a generalization of the identric mean. Indeed, it is immediately seen that µZ 1 ¶ I(x1 , x2 ) = exp ln(tx1 + (1 − t)x2 )dt , 0

in concordance with (5), which for µ = (n − 1)! gives the unweighted (and symmetric) identric mean of n arguments: Ã ! Z I = In = In (x1 , . . . , xn ) = exp (n − 1)!

En−1

ln(xλ)dλ1 . . . dλn−1

(6)

Let I 0 = In0 = In (1 − x) in (5) for µ = (n − 1)!. Then ndor ¸¶ and Trif µ Sáµ 1 . The [10] proved that relation (3) holds true for any n ≥ 2 xi ∈ 0, 2 weighted versions hold also true. In 1990 J. Sándor [7] discovered ¸ the following additive analogue of the Ky µ 1 (1 ≤ i ≤ n), then Fan inequality (1): If xi ∈ 0, 2 1 1 1 1 − ≤ 0− 0 H H A A

(7)

In 2002, E. Neuman and J. Sándor [2] proved the following refinement of (7): 1 1 1 1 1 1 − ≤ 0− ≤ 0− , (8) 0 H H L L A A where L is the (unweighted) logarithmic mean, obtained from (4) for f (t) = 1/t, i.e. Ã !−1 Z 1 L = Ln = Ln (x1 , . . . , xn ) = (n − 1)! dλ1 . . . dλn−1 , (9) En−1 xλ and L0 = L(1 − x). For n = 2 this gives the logarithmic mean of two arguments, L(x1 , x2 ) =

x2 − x1 ln x2 − ln x1

(x1 6= x2 ),

440

L(x, x) = 1.

We note that for n = 2, relation (8) is exactly inequality (27) in Rooin’s paper [6]. Alzer ([1]) proved another refinement of Sándor inequality, as follows: 1 1 1 1 1 1 − ≤ 0− ≤ 0− 0 H H G G A A

(10)

In [2] we have introduced a new mean J = Jn and deduced a new refinement of the Wang-Wang inequality: H J G ≤ 0 ≤ 0 0 H J G

(11)

We note that in a recent paper, Neuman and Sándor [4] have proved the following strong improvements of Alzer’s inequality (10): 1 1 1 1 1 1 1 1 1 1 ≤ 0− ≤ 0− ≤ 0− ≤ 0− − 0 H H J J G G I I A A

(12)

(where J 0 = J(1 − x) etc.).

2. New means and Ky Fan type inequalities 2.1. The results obtained by J. Rooin [6] are based essentially on the following Lemma 1. Let f be a convex function defined on a convex set C, and let xi ∈ C, 1 ≤ i ≤ n. Define F : [0, 1] → R by n

1X F (t) = f [(1 − t)xi + txn+1−i ], n

t ∈ [0, 1].

i=1

Then

µ

¶

f (x1 ) + · · · + f (xn ) , n Z 1 and the similar double inequality holds for F (t)dt. f

x1 + · · · + xn n

≤ F (t) ≤

0

Proof. By the definition of convexity, one has f [(1 − t)xi + txn+1−i ] ≤ (1 − t)f (xi ) + tf (xn+1−i ), and after summation, remarking that n X

[f (xn+1−i ) − f (xi )] = 0,

i=1

441

we get the right-side inequality. On the other hand, by Jensen’s discrete inequality for convex functions, ! Ã n µ ¶ x1 + · · · + xn 1X [(1 − t)xi + txn+1−i ] = f F (t) ≥ f , n n i=1

giving the left-side inequality. By integrating on [0, 1], clearly the same result holds true. 2.2. Now define the following mean of n arguments: Ã n !1/n Y K = Kn = Kn (x1 , . . . , xn ) = I(xi , xn+1−i ) (13) i=1

Letting f (x) = − ln x for x ∈ (0, +∞), and remarking that Z 1 ln[(1 − t)a + tb]dt = ln I(a, b), 0

Lemma 1 gives the following new refinement of the arithmetic-geometric inequality: G ≤ K ≤ A, (14) which holds true for any xi > 0 (i = 1,µn). ¸ 1−x 1 Selecting f (x) = ln for C = 0, , and remarking that x 2 Z 1 ln{1 − [(1 − t)a + tb]}dt = ln I(x01 , x02 ) = ln I 0 (x1 , x2 ), 0

we get the following Ky Fan-type inequality: G K A ≤ 0 ≤ 0 0 G K A

(15)

This is essentially inequality (13) in [6] (discovered independently by the author). 1 2.3. Let now f (x) = for x ∈ (0, ∞). Since f is convex, and x Z 1 1 1 dt = , (1 − t)a + tb L(a, b) 0 Lemma 1 gives H ≤ R ≤ A, 442

(16)

where R = Rn = Rn (x1 , . . . , xn ) = n/

n X i=1

1 L(xi , xn+1−i )

(17)

This is a refinement - involving the new mean R - of the harmonicarithmetic inequality. µ ¸ 1 1 1 Letting f (x) = − for x ∈ 0, , the above arguments imply the x 1−x 2 relations 1 1 1 1 1 1 − 0 ≤ − 0 ≤ − 0, (18) A A R R H H µ ¸ 1 where xi ∈ 0, , and R0 = Rn0 = Rn (1 − x1 , . . . , 1 − xn ). Relation (18) 2 coincides essentially with (26) of Rooin’s paper [6]. 2.4. Let S = Sn (x1 , . . . , xn ) = (xx1 1 . . . xxnn )1/(x1 +···+xn )

(19)

For n = 2, this mean has been extensively studied e.g. in [8], [9], [3]. Applying the Jensen inequality for the convex function f (x) = x ln x (x > 0), we get A ≤ S. On the other hand, remarking that S is a weighted geometric mean of x1 , . . . , xn with weights α1 = x1 /(x1 + · · · + xn ), . . . , αn = xn /(x1 + · · · + xn ), by applying the weighted geometric-arithmetic inequality xα1 1 . . . xαnn ≤ α1 x1 + · · · + αn xn , we can deduce S ≤ Q, where Q = Qn (x1 , . . . , xn ) =

x21 + · · · + x2n . x1 + · · · + xn

Therefore, we have proved that A≤S≤Q

(20)

In [8] it is shown that Z

b

x ln xdx = a

b2 − a2 ln I(a2 , b2 ) 4 443

(21)

Denote J(a, b) = (I(a2 , b2 ))1/2 and put J 0 (a, b) = J(1 − a, 1 − b). By applying Lemma 1, we get A ≤ T ≤ S, (22) where the mean T is defined by #1/A " n Y A(xi ,xn+1−i ) n (J(xi , xn+1−i )) T = Tn (x1 , . . . , xn ) =

(23)

i=1

µ ¸ 1 Letting now f (x) = x ln x − (1 − x) ln(1 − x), x ∈ 0, , by f 00 (x) = 2 1 − 2x ≥ 0 we can state that f is convex, so by Lemma 1 and by (21) we x(1 − x) ¸ µ 1 can write, for xi ∈ 0, : 2 AA /A0

A0

≤ T A /T 0

A0

A0

≤ S A /S 0 ,

(24)

where the mean T is defined by (23), while T 0 = T (1−x). Since for n = 2, T ≡ J, for means of two arguments (24) gives a Ky Fan-type inequality involving A, I, S. 2.5. Relation (23) shows that T is a generalization of the mean J to n arguments. In what follows we shall introduce another generalization, provided by the formula U = Un (x1 , . . . , xn ) ( Ã !)1/A Z = exp (n − 1)! (xλ) ln(xλ)dλ1 . . . dλn−1 (25) En−1

Here the notations are as in the Introduction. Since, by (21), Z b Z 1 1 [(1 − t)a + tb] ln[(1 − t)a + tb]dt = x ln xdx b−a a 0 A ln I(a2 , b2 ) = ln J A , 2 for n = 2, we have U ≡ J, thus U is indeed another generalization of the mean J. Now, the following result is due to E. Neuman (see e.g. [2]). Lemma 2. Let K be an interval containing x1 , . . . , xn , and suppose that f : K → R is convex. Then ¶ µ Z x1 + · · · + xn ≤ (n − 1)! f (λx)dλ1 . . . dλn−1 f n En−1 =

444

≤

f (x1 ) + · · · + f (xn ) . n

µ ¸ 1 Letting K = 0, , and f (x) = x ln x − (1 − x) ln(1 − x) in Lemma 2, we can µ2 ¸ 1 deduce for xi ∈ 0, 2 AA /A0

A0

≤ U A /U 0

A0

≤ S A /S 0

A0

(26)

Remark that for n = 2, inequalities (24) and (26) reduce to the same inequality, as in that case one has T = J = U . The mean U separates also A and S, since applying Lemma 2 for f (x) = x ln x (x > 0), we have A ≤ U ≤ S.

(27)

There remains an Open Problem, namely the comparability of the above defined means T and U for n > 2. Also, the connections of these means to K and R, introduced in the preceding sections.

References [1] H. Alzer, The inequality of Ky Fan and related results, Acta Appl. Math. 38, 305-354(1995). [2] E. Neuman, J. Sándor, On the Ky Fan inequality and related inequalities, I, Math. Ineq. Appl. 5, 49-56(2002). [3] E. Neuman, J. Sándor, Inequalities involving Stolarsky and Gini means, Math. Pannonica 14, 29-44(2003). [4] E. Neuman, J. Sándor, On the Ky Fan inequality and related inequalities, II, submitted. [5] A.O. Pittenger, The logarithmic mean in n variables, Amer. Math. Monthly 92, 99-104(1985). [6] J. Rooin, Some refinements of Ky Fan’s and S´ andor’s inequalities, Southest Asian Bull. Math. 27, 1101-1109(2004). [7] J. Sándor, On an inequality of Ky Fan, Babe¸s-Bolyai Univ., Fac. Math. Phys., Res. Semin. 7, 29-34(1990). 445

[8] J. Sándor, On the identric and logarithmic means, Aequationes Math. 40, 261-270(1990). [9] J. Sándor, I. Ra¸sa, Inequalities for certain means of two arguments, Nieuw Arch. Wiskunde 15, 51-55(1997). [10] J. Sándor, T. Trif, A new refinement of the Ky Fan inequality, Math. Ineq. Appl. 4, 529-533(1999). [11] W.-L. Wang, P.-F. Wang, A class of inequalities for symmetric functions (in Chinese), Acta Math. Sinica 27, 485-497(1984).

10

On Lehman’s inequality and electrical networks

1. Introduction A. Lehman’s inequality (see [6], [2]) (and also SIAM Review 4(1962), 150155), states that if A, B, C, D are positive numbers, then (A + B)(C + D) AC BD ≥ + . (1) A+B+C +D A+C B+D This was discovered as follows: interpret A, B, C, D as resistances of an electrical network. It is well-known that if two resistances R1 and R2 are serially connected, then their compound resistance is R = R1 + R2 , while in parallel connecting one has 1/R = 1/R1 + 1/R2 . Now consider two networks, as given in the following two figures: (A + B)(C + D) AC BD R0 = + A+B+C +D A+C B+D By Maxwell’s principle, the current chooses a distribution such as to minimize the energy (or power), so clearly R0 ≤ R, i.e. Lehman’s inequality (1). In fact, the above construction may be repeated with 2n resistances, in order to obtain: Theorem 1. If ai , bi (i = 1, n) are positive numbers, then R=

a1 b1 an bn (a1 + · · · + an )(b1 + · · · + bn ) ≥ + ··· + a1 + · · · + an + b1 + · · · + bn a1 + b1 an + bn

(2)

for any n ≥ 2.

2ab = H(a, b) is in fact the harmonic mean of two a+b positive numbers, Lehman’s inequality (2) can be written also as Remark. Since

H(a1 + · · · + an , b1 + · · · + bn ) ≥ H(a1 , b1 ) + · · · + H(an , bn ) 446

(3)

A

C

A

C

B

D

B

D

2. Two variable generalizations In what follows, by using convexity methods, we shall extend (3) in various ways. First we introduce certain definitions. Let f : A ⊂ R2 → R be a function with two arguments, where A is a cone (e.g. A = R2+ ). Let k ∈ R be a real number. Then we say that f is k-homogeneous, if f (rx, ry) = rk f (x, y)

(4)

for any r > 0 and x, y ∈ A. When k = 1, we simply say that f is homogeneous. Let F : I ⊂ R → R be a function of an argument defined on an interval I. We say that F is k-convex (k-concave), if F (λa + µb) ≤ λk F (a) + µk f (b),

(5)

(≥)

for any a, b ∈ I, and any λ, µ > 0, λ + µ = 1. We note, that if k = 1, then F will be called simply convex. For example, F (t) = |t|k , t ∈ R is k-convex, for k ≥ 1, since |λa+µb|k ≤ λk |a|k +µk |b|k by (u+v)k ≤ uk +v k (u, v > 0), k ≥ 1, which is well-known. On the other hand, the function F (t) = |t|, though is convex, is not 2-convex on R. The k-convex functions have been introduced for the first time by W. W. Breckner [4]. See also [5] for other examples and results. A similar convexity notion, when in (5) one replaces λ + µ = 1 by λk + µk = 1, was introduced by W. Orlicz [12] (see also [8] for these convexities). 447

Now, let A = (0, +∞) × (0, +∞) = R2+ and I = (0, +∞). Define F (t) = f (1, t) for t ∈ I. Theorem 2. If f is k-homogeneous, and F is k-convex (k-concave) then f (a1 + · · · + an , b1 + · · · + bn ) ≤ f (a1 , b1 ) + · · · + f (an , bn )

(6)

(≥)

for any ai , bi ∈ A (i = 1, 2, . . . , n). Proof. First remark, that by (4) and the definition of F , one has µ ¶ µ ¶ b b ak F = ak f 1, = f (a, b) a a

(7)

On the other hand, by induction it can be proved the following Jensen-type inequality: F (λ1 x1 + λ2 x2 + · · · + λn xn ) ≤ λk1 F (x1 ) + λk2 F (x2 ) + · · · + λkn F (xn ),

(8)

(≥)

for any xi ∈ I, λi > 0 (i = 1, n), λ1 + · · · + λn = 1. E.g. for n = 3, relation (8) can be proved as follows: λ1 λ2 Put a = x1 + x2 , b = x3 , λ = λ1 + λ2 , µ = λ3 in (5). Then, λ1 + λ2 λ1 + λ2 as λ1 x1 + λ2 x2 + λ3 x3 = λa + µb, we have F (λ1 x1 + λ2 x2 + λ3 x3 ) ≤ λk F (a) + µk F (b) ≤ · k

≤ (λ1 + λ2 )

¸ λk1 λk2 F (x1 ) + F (x2 ) + λk3 F (x3 ) = (λ1 + λ2 )k (λ1 + λ2 )k = λk1 F (x1 ) + λk2 F (x2 ) + λk3 F (x3 ).

Put now in (8) x1 = λ1 =

b1 b2 bn , x2 = , . . . , xn = , a1 a2 an

a1 a2 an , λ2 = , . . . , λn = a1 + · · · + an a1 + · · · + an a1 + · · · + an

in order to obtain µ µ F

b1 + · · · + bn a1 + · · · + an

ak1 F

¶ ≤ (≥)

b1 a1

¶

µ

µ ¶ ¶ bn b2 k + + · · · + an F a2 an (a1 + · · · + an )k ak2 F

448

(9)

Now, by (7) this gives f (a1 + · · · + an , b1 + · · · + bn ) ≤ f (a1 , b1 ) + · · · + f (an , bn ), i.e. relation (6). a+b Remark. Let f (a, b) = . Then f is homogeneous (i.e. k = 1), and ab t+1 F (t) = f (1, t) = is 1-convex (i.e., convex), since F 00 (t) = 2/t3 > 0. Then t relation (6) gives the following inequality: 1 1 1 ≤ + ··· + . H(a1 + · · · + an , b1 + · · · + bn ) H(a1 , b1 ) H(an , bn )

(10)

ab t . Then f is homogeneous, with F (t) = , which a+b t+1 is concave. From (6) (with ≥ inequality), we recapture Lehman’s inequality (3). The following theorem has a similar proof: Theorem 3. Let f be k-homogeneous, and suppose that F is l-convex (l-concave) (k, l ∈ R). Then Let now f (a, b) =

(a1 + · · · + an )l−k f (a1 + · · · + an , b1 + · · · + bn ) ≤ (≥) l−k al−k 1 f (a1 , b1 ) + · · · + an f (an , bn ).

(11)

Remarks. For k = l, (11) gives (9). a For example, let f (a, b) = , where a, b ∈ (0, ∞) × (0, ∞). Then k = 0 b 1 (i.e. f is homogeneous of order 0), and F (t) = , which is 1-convex, since t 2 00 F (t) = 3 > 0. Thus l = 1, and relation (11) gives the inequality t (a1 + · · · + an )2 a2 a2 ≤ 1 + ··· + n b1 + · · · + bn b1 bn Finally, we given another example of this type. Put f (a, b) =

(12) a2 + b2 . a+b

t2 + 1 , after elementary computations, F 00 (t) = t+1 4/(t + 1)3 > 0, so l = 1, and (11) (or (9)) gives the relation Then k = 1. Since F (t) =

a2 + b21 a2 + b2n (a1 + · · · + an )2 + (b1 + · · · + bn )2 ≤ 1 + ··· + n a1 + · · · + an + b1 + · · · + bn a1 + b1 an + bn 449

(13)

a2 + b2 ap+1 + bp+1 (more generally, Lp (a, b) = ) are the a+b ap + bp so-called ”Lehmer means” [9], [7], [1] of a, b > 0, (13) can be written also as Since L1 (a, b) =

L1 (a1 + · · · + an , b1 + · · · + bn ) ≤ L1 (a1 , b1 ) + · · · + L1 (an , b1 ).

(14)

Clearly, one can obtain more general forms for Lp . For inequalities on more general means (e.g. Gini means), see [10], [11].

3. H¨ older’s inequality As we have seen, there are many applications to Theorems 2 and 3. Here we wish to give an important application; namely a new proof of Hölder’s inequality (one of the most important inequalities in Mathematics). Let f (a, b) = a1/p b1/q , where 1/p + 1/q = 1 (p > 1). Then clearly f 1 is homogeneous (k = 1), with F (t) = t1/q . Since F 0 (t) = t−1/p , F 00 (t) = q 1 −(1/p)−1 − t < 0, so by Theorem 2 one gets pq 1/p 1/q

(a1 + · · · + an )1/p (b1 + · · · + bn )1/q ≥ a1 b1

1/q + · · · + a1/p n bn

(15)

Replace now ai = Api , bi = Biq (i = 1, n) in order to get n X i=1

Ai Bi ≤

Ã n X

Api

!1/p Ã n X

!1/q Biq

,

(16)

i=1

i=1

which is the classical Hölder inequality.

4. Many variables generalization Let f : A ⊂ Rn+ → R be of n arguments (n ≥ 2). For simplicity, put p = (x1 , . . . , xn ), p0 = (x01 , . . . , x0n ), when p + p0 = (x1 + x01 , . . . , xn + x0n ) and rp = (rx1 , . . . , rxn ) for r ∈ R. Then the definitions of k-homogeneity and k-convexity can be extended to this case, similarly to paragraph 2. If A is a cone, then f is k-homogeneous, if f (rp) = rk f (p) (r > 0) and if A is convex set then f is k-convex, if f (λp + µp0 ) ≤ λk f (p) + µk f (p0 ) for 0 any µ p, p 0 ∈ ¶ A, λ, µ > 0, 0λ + µ = 1. We say that f is k-Jensen convex, if p+p f (p) + f (p ) f ≤ . We say that f is r-subhomogeneous of order 2 2k k, if f (rp) ≤ rk f (p). Particularly, if k = 1 (i.e. f (rp) ≤ rf (p)), we say that f is r-subhomogeneous (see e.g. [14], [15]). If f is r-subhomogeneous of order 450

k for any r > 1, we say that f is subhomogeneous of order k. For k = 1, see [13]. We say that f is subadditive on A, if f (p + p0 ) ≤ f (p) + f (p0 )

(17)

We note that in the particular case of n = 2, inequality (6) with ”≤” says exactly that f (a, b) of two arguments is subadditive. Theorem 4. If f is homogeneous of order k, then f is subadditive if and only if it is k-Jensen convex. Proof. If f is subadditive, i.e. f (p + p0 ) ≤ f (p) + f (p0 ) for any p, p0 ∈ A, then µ ¶ p + p0 1 f (p) + f (p0 ) f = k f (p + p0 ) ≤ , 2 2 2k so f is k-Jensen convex. Reciprocally, if f is k-Jensen convex, then µ ¶ p + p0 f (p) + f (p0 ) f ≤ , 2 2k so

· µ ¶¸ µ ¶ p + p0 p + p0 f (p + p0 ) = f 2 = 2k f ≤ f (p) + f (p0 ), 2 2

i.e. (17) follows. Remark. Particularly, a homogeneous subadditive function is convex, a simple, but very useful result in the theory of convex bodies (e.g. ”distance function”, ”supporting function”, see e.g. [3], [16]). Theorem 5. If f is 2-subhomogeneous of order k, and is k-Jensen convex, then it is subadditive. Proof. Since µ µ ¶¶ µ ¶ p + p0 p + p0 f (p + p0 ) = f 2 ≤ 2k f , 2 2 and

µ f

p + p0 2

¶ ≤

f (p) + f (p0 ) , 2k

we get f (p + p0 ) ≤ f (p) + f (p0 ), so (17) follows. Remark. Particularly, if f is 2-subhomogeneous, and Jensen convex, then it is subadditive. (18) It is well-known that a continuous Jensen convex function (defined on an open convex set A ⊂ Rn ) is convex. Similarly, for continuous k-Jensen convex functions, see [4]. 451

To give an interesting example, connected with Lehman’s inequality, let us ¶ µ 1 1 + ··· + . consider A = Rn+ , f (p) = H(p) = n/ x1 xn 1 1 1 Let = + ··· + . Then g(p) x1 xn n

dg X dxi = , g2 x2i i=1

so 1 d2 g = 2 g3

Ã

n X dxi i=1

x2i

n

X dx2 dg 2 d2 g i − 2 = −2 , g2 g3 x3i i=1

!2

Ã −

n X 1 xi i=1

!Ã

n X dx2 i

i=1

x3i

! .

(Here d denotes a differential.) Now apply Hölder’s inequality (16) for p = q = √ √ 2 (i.e. Cauchy-Bunjakovski inequality), Ai = 1/ xi , Bi = (1/xi xi )dxi . Then d2 g one obtains 3 ≤ 0, and since g > 0, we get d2 g ≤ 0. It is well-known ([16]) g that this implies the concavity of function g(p) = H(p)/n, so −H(p) will be a convex function. By consequence (17) of Theorem 5, H(p) is subadditive, i.e. H(x1 + x01 , x2 + x02 , . . . , xn + x0n ) ≥ H(x1 , x2 , . . . , xn )+ +H(x01 , x02 , . . . , x0n ),

(xi , x0i > 0).

(19)

For n = 2 this coincides with (3), i.e. Lehman’s inequality (1). Finally, we prove a result, which is a sort of reciprocal to Theorem 5: Theorem 6. Let us suppose that f is subadditive, and k-convex, where k ≥ 1. Then f is subhomogeneous of order k. Proof. For any r > 1 one can find a positive integer n such that r ∈ [n, n + 1]. Then r can be written as a convex combination of n and n + 1: r = nλ + (n + 1)µ. By the k-convexity of f one has f (rp) = f (nλp + (n + 1)µp) ≤ λk f (np) + µk f [(n + 1)p]. Since f is subadditive, from (17) it follows by induction that f (np) ≤ nf (p), so we get f (rp) ≤ nλk f (p) + (n + 1)µk f (p) = [nλk + (n + 1)µk ]f (p). Now, since k ≥ 1, it is well-known that [λn + (n + 1)µ]k ≥ (λn)k + ((n + 1)µ)k . 452

But (λn)k ≥ nλk and ((n + 1)µ)k ≥ (n + 1)µk , so finally we can write f (rp) ≤ [λn + (n + 1)µ]k f (p) = rk f (p), which means that f is subhomogeneous of order k. Remark. For k = 1 Theorem 6 contains a result by R. A. Rosenbaum [13]. Acknowledgments. The author thanks Professors J. Peetre and H. Alzer for providing their reprints [2], resp. [1]. He is indebted to Professor F. A. Valentine for a copy of his book [16], and also to Professors W. W. Breckner of Cluj, and V. E. Szabó of Budapest for helpful discussions. NOTE ADDED IN PROOF. Recently (23th February, 2005) we have discovered that Lehman’s inequality (2) (or (3)) appears also as Theorem 67 in G. H. Hardy, J. E. Littlewood and G. Polya [Inequalities, Cambridge Univ. Press, 1964; see p.61], and is due to E. A. Milne [Note on Rosseland’s integral for the stellar absorption coefficient, Monthly Notices, R.A.S. 85(1925), 979-984]. Though we are unable to read Milne’s paper, perhaps we should call Lehman’s inequality as the ”Milne-Lehman inequality”. We note also that the Milne-Lehman inequality is published as a Proposed Problem 2113 (by M.E. Kuczma), as well as Problem 2392 (by G. Tsintsifas) in the Canadian journal Crux Mathematicorum.

References ¨ Lehmers Mittelwertfamilie, Elem. Math., 43(1988), no. [1] H. Alzer, Uber 2, 50-54. [2] J. Arazy, T. Claesson, S. Janson and J. Peetre, Means and their iteration, Proc. 19th Nordic Congr. Math., Reykjavik 1984, 191-212. [3] T. Bonnesen and W. Fenchel, Theorie der konvexen K¨ orper, Berlin, 1934. [4] W. W. Breckner, Stetigkeitsaussagen f¨ ur eine Klasse verallgemeinerter konvexer Funktionen in topologischen linearen R¨ aume, Publ. Inst. Math. (Beograd), 23(1978), 13-20. [5] W. W. Breckner and Gh. Orbán, Continuity properties of rationally sconvex mappings with values in an ordered topological linear space, 92 pp., Babe¸s-Bolyai Univ., Cluj, Romania, 1978. [6] R. J. Duffin, Network models, in SIAM-AMS Proceedings vol. III, pp. 65-91, AMS, Providence, 1971. 453

[7] H. W. Gould and M. E. Mays, Series expansions of means, J. Math. Anal. Appl., 101(1984), 611-621. [8] H. Hudzik and L. Maligranda, Some remarks on s-convex functions, Aequationes Math., 48(1994), 100-111. [9] D. H. Lehmer, On the compounding of certain means, J. Math. Anal. Appl., 36(1971), 183-200. [10] E. Neuman and J. Sándor, Inequalities involving Stolarsky and Gini means, Math. Pannonica 14(2003), 29-44. [11] E. Neuman and J. Sándor, On certain new means of two arguments and their extensions, Int. J. Math. Math. Sci., 16(2003), 981-993. [12] W. Orlich, A note on modular spaces, I, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys., 9(1961), 157-162. [13] R. A. Rosenbaum, Subadditive functions, Duke Math. J., 17(1950), 227247. [14] J. Sándor and Gh. Toader, On some exponential means, Babe¸s-Bolyai Univ., Preprint No. 7, 1990, 35-40. [15] J. Sándor, On certain subhomogeneous means, Octogon Math. Mag., 8(2000), 156-160. [16] F. A. Valentine, Convex sets, Mc Graw-Hill Inc., New York, 1964.

454

Author Index A ˇ Arslanagić 1.4, 4.4 S. H. Alzer 2.1, 2.10, 4.9, 4.10, 4.16, 5.12, 6.3, 6.4, 6.10, 6.13, 6.14, 6.15, 10.9, 10.10 J.L. d’Alambert 2.5 T.M. Apostol 3.7, 9.10 Ch. Ashbacher 3.11, 3.12, 3.17, 9.16 W. Aiello 3.19 E. Artin 5.9, 5.11, 5.13, 5.15, 6.11, 7.1 R.P. Agarwal 5.13 M. Abramowitz 5.13 C. Alsina 5.15 J. Arazy 6.3, 10.10 R. Askey 7.1 J. Aczél 7.2, 7.4, 7.5 D. Acu 7.2 T. Andreescu 8.3 D. Andrica 8.3 C. Adiga 9.4 L. Alaoglu 9.7 K.T. Atanassov 9.14, 10.8

O. Bottema 1.5, 1.6 H. Brocard 1.6 D.M. B˘atinet¸u-Giurgiu 1.8, 2.2A, 2.2L, 2.2M, 2.2N, 2,2O, 2.2P, 2.2S, 2.2T, 5.5, 5.6, 5.7 H.J. Brothers 2.1, 2.2F, 2.2P J. Bernoulli 2.2I, 2.2J, 3.7, 5.3 V. Berinde 2.5 T.J. Bromwich 2.5 Gy. Bereznai 2.6 N.G. de Bruijn 2.7 F. Bencherif 2.7 D. Bradley 2.9 G.M. Bell 2.9 T. Bonnesen 10.10 G. Bennett 2.10, 5.12 A. Balog 3.2 D. Bode 3.9 R. Bojanić 3.16 A. Bege 3.19 V. Bunjakovski (Bunyakovsky) 4.2, 4.4, 6.6, 10.10 E.F. Beckenbach 4.14 R. Bellman 4.14 B H. Bohr 5.9, 5.13, 7.1 E. Bézout 1.2 J. Bass 5.11 M. Bencze 1.5, 1.8, 1.11, 1.12, 2.2A, J. Bursuc 5.11 2.2B, 2.2C, 2.2G, 2.2H, 2.2J, 2.2S, N.S. Barnett 5.13 2.2T, 2.2U, 2.3, 2.9, 3.9, 3.10, 3.14, C.W. Borchardt 6.7 4.13, 5.5, 5.6, 5.14, 8.3, 8.9, 9.6, 9.9 455

O. Bonnett 6.16 M. Balázs 6.16 V. B˘andil˘a 7.2 B. Bartha 7.9 A. Bremner 8.6 J.M. Brown 10.3 W.W. Breckner 10.10

L. Denbath 2.2P G.L. Dirichlet 2.9, 9.7, 9.14 S.S. Dragomir 2.10, 4.9, 4.10, 5.13 P. Dusart 3.2 F.W. Dodd 3.8 R. Dedekind 3.13, 8.12, 8.13, 8.14 L.E. Dickson 3.13 M. Deng 3.19 C B.P. Demidovich 5.11, 6.11 R. Cotes 1.2 B.C. Drachman 5.16 L. Cauchy 1.5, 2.2H, 2.10, 4.2, 4.4, J. Dieudonné 6.14 4.16, 6.6, 6.10, 6.13, 6.15, 7.1, 7.3, 7.4, G. Darboux 6.15, 6.16 7.5, 7.6, 10.10 R.J. Duffin 10.10 L. Carlitz 1.6 F.T. Câmpan 1.14 E N. de Cusa 1.14 P. Erdös 1.3, 1.9. 9.3, 9.7, 9.14 E. Cesarò 2.2I, 2.2O, 2.2R, 2.3A, 2.3B L. Euler 1.5, 2.2J, 2.2K, 2.2M, 2.2O, J.H. Conway 2.2U 2.2Q, 2.3, 2.6, 2.9, 2.10, 3.7, 3.11, 3.12, J. Choi 2.9 3.13, 5.1, 5.2, 5.4, 5.5, 5.7, 5.11, 6.11, E. Catalan 2.9 6.16, 8.11, 8.12, 8.13, 9.7, 9.8, 9.14, B. Crstici 3.1, 3.2, 5.15, 9.1, 9.2, 10.2 10.4 W. Chen 3.8 A. Erdélyi 2.4 H.Z. Cao 3.8 Euclid of Alexandria 3.12, 10.2 G.L. Cohen 3.19, 10.7 J. Earls 3.15, 3.16, 3.17, 3.18, 8.16, G. Chrystal 4.3, 4.4 10.5 P. Chebyshev 4.10, 4.11, 7.10 F M. Craiu 5.11, 6.16 P. Fermat 1.2, 3.17, 10.2 M.J. Cloud 5.16 B. Finta 1.3 B.C. Carlson 6.3, 6.7, 6.11 J.B. Fourier 1.11 T. Claeson 6.7, 10.10 S. Finch 2.1 C.-P. Chen 6.13 L. Filep 2.6 G. Chilov 6.16 K. Ford 3.15 C.V. Cr˘aciun 6.16 J. Findley 3.15 R.O. Cuzmin 6.16 J. Fabrykowski 3.19 N. Cior˘anescu 7.4 A.M. Fink 4.12 B. Conn 8.6 K. Fan 4.13, 4.14, 6.10, 10.9 E. Cohen 9.6, 9.10, 9.11 J. Frauenthal 5.16 I. Creang˘a 9.7 A.N. Fathima 9.4 D W. Fenchel 10.10 456

G R. Ianić 7.2 M. Gherm˘anescu 2.1, 2.2O, 2.2P J G. Garnir 2.2I, 2.2R C. Jordan 1.4, 1.11, 6.7, 9.7 R.K. Guy 3.4, 8.1, 8.6, 8.15, 10.7 J.L. Jensen 1.11, 4.7, 4.9, 5.13, 7.4, P. Gronas 3.12 K.F. Gauss 3.12, 3.13, 5.4, 6.3, 6.7, 9.11, 9.13, 10.9, 10.10 W. Janous 1.12, 3.1 9.6, 10.2 S. Janson 6.7, 10.10 S.W. Golomb 3.13 F.H. Jackson 7.1 J. Galambos 3.14 W. Gautschi 5.13 K B. Gelbaum 6.14 N.D. Kazarinoff 1.9 N.M. G¨ unther 6.16 J.A. Knox 2.1, 2.2F, 2.2O R.R. Goldberg 6.16 K. Knopp 2.8 V.L. Gardiner 8.6 H.M. Korobov 3.4 C. Gini 10.9, 10.10 H.J. Kanold 3.9, 3.10 H.W. Gould 10.10 I. Kátai 3.14 J.M. DeKoninck 3.19, 9.6, 9.10 H N. Keyfitz 5.16 Heron of Alexandria J. Hadamard 1.11, 2.1, 4.7, 4.8, 4.9, F. Klein 6.4 J. Karamata 6.5 4.10, 5.11, 5.13, 6.9, 9.11, 9.13 J. Kolumb´ an 6.16 Ch. Hermite 1.11, 4.7, 5.13 K. Koyama 8.6 G. de l’Hôpital 1.15, 2.1, 2.2R T. Kim 9.4 D.R. Hofstadter 2.2U G.H. Hardy 2.9, 2.10, 3.9, 4.3, 4.11, M.S. Klamkin 9.11, 10.3 4.14, 5.13, 6.10, 6.14, 8.1, 9.6, 9.7, K. Kashihara 9.16 E. Krätzel 10.3 9.10, 9.11, 10.3 J. Van der Hoek 2.10 L M. Hazewinkel 3.14 P.S. Laplace 6.16 P. Hagis 3.19 T. Lalescu 1.6, 5.1, 5.12 J. Hanumanthachari 3.19 F. Leuenberger 1.6 O. Hölder 5.13, 5.14, 7.1, 10.10 E. Lemoine 1.6 V. Hiri¸s 6.16 J. Lagrange 2.2Q, 2.10, 4.9, 5.2, 6.7, S. Haruki 7.4 6.9, 6.11 D.R. Heath-Brown 8.6 D.A. Lavis 2.9 M. Hausman 9.2 J.E. Littlewood 2.10, 3.4, 4.3, 4.11, H. Hudzik 10.10 4.14, 5.13, 6.10, 6.14 I G.G. Lorentz 2.10 A. Ivić 3.19, 9.6 G. Lord 3.19 457

L. Lucht 3.19 T.P. Lin 4.7, 6.3 A.M. Legendre 5.4 A. Lupa¸s 5.11 J. Lew 5.16 A. Lehman 10.10 D.H. Lehmer 10.10 E.B. Leach 6.2, 6.3, 6.5, 6.7 H. Lebesgue 7.1 F. Luca 9.14, 10.8 C.C. Lindner 10.8 E.S. Langford 9.11 B. Lepson 10.3 W.M. Lioen 8.6 R.F. Lukes 8.6 R.B. Lazarus 8.6 R. Lipschitz 9.1 M. Lee 9.2, 9.14, 10.8 R. Laatsch 9.2 E. Lucas 9.8

L. Maligranda 10.10 I.A. Maron 5.11, 6.11 M. Megan 6.16 J.C.P. Miller 8.6 L.J. Mordell 8.6 P. Moree 9.3, 9.14, 9.15, 10.8 A. Möbius 9.7 A. Makowski 9.10 J. Morgado 9.10 M.E. Mays 10.10 J.C. Maxwell 10.10

M D.S. Mitrinović 1.4, 1.5, 1.11, 1.13, 1.16, 2.10, 3.1, 3.2, 3.4, 3.8, 4.10, 4.12, 5.8, 5.10, 5.11, 5.15, 6.11, 7.1, 9.1, 9.2, 9.6, 9.11, 10.3 D.M. Milosević 1.4, 1.5, 4.10 H. Minc 2.10, 5.11 R.K. Meany 10.3 Gy. Maurer 1.2 K.B. Mollweide 1.7 L.J. Mordell 1.9 A.W. Marshall 1.12 C. Maclaurin 2.2H, 2.3A, 6.10 A. Murthy 2.8, 3.15, 9.2, 9.3, 9.6 J.S. Martins 2.10 L.E. Matties 3.8 A.P. Minin 3.13 P. Mersenne 3.15 J. Mollerup 4.9, 5.13, 7.1

N C. Nedelcu 1.4 T. Negoi 2.2F J. Nagura 3.5 C.A. Nicol 3.13, 8.11 K. Nageswara Rao 3.19 E. Neuman 4.13, 10.9, 10.10 S.M. Nikolsky 6.16 T. Nagell 8.4 A. Nikiforov 5.15 O M. Olteanu 1.10 I. Olkin 1.12 L. Olivier 2.8 O. Ore 3.19, 10.7 J.M.H. Olmsted 6.14 W. Orlicz 10.10 Gh. Orbán 10.10 V. Ouvarov 5.15 P J.E. Peˇcarić 1.5, 4.9, 4.10, 7.2 G. Pólya 1.6, 2.10, 4.3, 4.11, 4.14, 5.13, 6.10, 6.14, 9.7 D. Pedoe 1.6 A.O. Pittenger 2.1, 6.3, 10.9 T. Popoviciu 2.1

458

ˇ I.I. Pjateckii-Sapiro 3.2 F. Popovici 3.9 J.L. Pe 3.12, 3.13 M. Petrović 5.9 J. Peetre 6.7, 10.10 C. Popa 6.16 E. Prouchet 9.8 C. Pomerance 9.10 Q F. Qi 6.13 R R. Redheffer 1.4, 1.11, 1.13 G. Robin 2.7 S. Ramanujan 2.9, 6.5, 9.6 J.B. Rosser 3.1, 3.2, 3.6 B. Riemann 3.4, 3.7, 5.5, 6.16, 9.7 I.Z. Ruzsa 3.10, 9.11 A.W. Roberts 4.3, 4.15, 6.15 J.L. Raabe 5.11 I. Ra¸sa 6.3, 6.6, 10.9 M. Rolle 6.11, 6.14, 6.15 W. Rudin 6.14 M.N. Ro¸sculet¸ 5.11, 6.16 Th. M. Rassias 7.8, 7.12, 8.2, 8.5, 8.6 H.J.J. te Riele 8.6 H. Roskam 9.14, 9.15, 10.8 J. Rooin 10.9 R.A. Rosenbaum 10.10 P. du Bois-Reymond 6.16 S J. Sándor 1.3, 1.4, 1.5, 1.6, 1.8, 1.9, 1.10, 1.11, 1.13, 1.14, 2.1, 2.2A, 2.2B, 2.2C, 2.2D, 2.2F, 2.2H, 2.2J, 2.2O, 2.2P, 2.2S, 2.5, 2.9, 2.10, 3.1, 3.2, 3.4, 3.6, 3.7, 3.8, 3.9, 3.10, 3.11, 3.12, 3.13, 3.14, 3.16, 3.19, 4.3, 4.7, 4.8, 4.9, 4.10, 4.11, 4.15, 4.16, 5.5, 5.7, 5.8, 5.10,

5.11, 5.13, 5.14, 5.15, 6.1, 6.2, 6.3, 6.4, 6.6, 6.7, 6.8, 6.9, 6.10, 6.11, 6.13, 6.14, 6.15, 6.16, 7.2, 7.4, 7.5, 7.11, 8.1, 8.6, 8.9, 8.11, 8.12, 8.14, 8.15, 9.1, 9.3, 9.4, 9.5, 9.7, 9.10, 9.11, 9.13, 9.14, 9.15, 9.16, 10.2, 10.3, 10.4, 10.7 G. Szegö 1.6, 9.7 S.B. Steˇckin 1.11 W. Snellius 1.14 K.B. Stolarsky 2.1, 4.9, 6.3, 6.7, 10.3, 10.10 Gh. Stoica 2.1 O. Stolz 2.2I, 2.2O, 2.2Q, 2.3A, 2.3B I. Sathre 2.10, 5.11 I. Schoenfeld 3.1, 3.2, 3.6 J. Stirling 3.7, 5.5 D. Suryanarayana 3.9, 3.10 I. Smarandache 3.9, 3.11, 3.12, 3.15, 3.17, 8.1, 8.13, 8.16, 9.4, 9.5, 9.13, 9.14, 9.15, 9.16, 10.5 I. A. Stegun 5.13 R.M. Sorli 3.19, 10.7 E.G. Straus 3.19, 9.13 M.V. Subbarao 3.19, 9.13 H.J. Seiffert 6.1, 6.3, 6.4, 6.8 V.E.S. Szabó 6.1, 6.2, 6.3 M.C. Sholander 6.2, 6.3, 6.5, 6.7 J. Steiner 6.4 H. Shniad 6.5 W. Sierpinski 6.10, 7.3, 9.3, 9.9 E.L. Stark 6.16 I. Stamate 7.2 H.N. Shapiro 7.3, 9.8 K.B. Subramanian 8.4 P.R. Stein 8.6 H. Sekigawa 8.6 D.D. Somashekara 9.4 A. Schinzel 9.10, 9.14 A. Stenger 10.3

459

T F.A. Valentine 10.10 D.W. De Tempe 2.2F S.R. Tims 2.2F W J.A. Tyrrel 2.2F R. Weitzenb¨ ock F. Tricomi 2.4 J. Wallis 2.4 L. Tóth 2.4, 5.12, 9.10, 9.11 E.M. Wright 3.9, 8.1, 9.7, 9.10, 9.11, B. Taylor 3.7, 6.11, 6.15, 6.16 10.3 Gh. Toader 4.9, 4.10, 4.11, 6.3, 6.6, Ch. Wall 3.19 6.9, 10.10 E.T. Whittaker 5.3, 5.4, 5.5, 5.6, 5.11, T. Trif 4.13, 10.9 5.14, 5.15, 5.16, 6.16 M.S. Tomás 5.15 G.N. Watson 5.3, 5.4, 5.5, 5.6, 5.11, J. Thomae 7.1 5.14, 5.15, 5.16, 6.16 Y. Tsuruoka 8.6 K. Weierstrass 5.11, 6.16 Z. Tuzson 8.6 J. Wendel 5.13, 5.14, 5.16 E.C. Titchmarsh 9.7 D. Widder 6.16 R. Webster 7.1 V D.T.Walker 8.3, 8.4 I. Virág 1.2 M.F.C. Woollett 8.6 V. Volenec 1.5 W.-L. Wang 10.9 A. Vernescu 2.4 P.-F. Wang 10.9 Ch.J. de la Vallée-Poussin 3.4 P. Vlamos 3.6 Y D.E. Varberg 4.3, 4.15, 6.15 W.H. Young 5.10 M. Vuorinen 5.14, 6.2, 6.3, 6.7 M.K. Vamanamurthy 6.2, 6.3, 6.7 Z M. Vlada 7.2 A. Zygmund 6.16 I.M. Vinogradov 8.3 P. Zwier 10.3 L. Vaserstein 8.6

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